ARITHMETIC PROGRESSIONS

EXERCISE 5.1

Note: If a, b, c are in AP, then b = (a + c)/2 and b is called the arithmetic mean of a and c.

1. In which of the following situations, does the list of numbers involved make an arithmetic

progression, and why?

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each

additional km.

Solution:

The taxi fare is ₹15 for the first km and ₹8 for each additional km.

Let’s list the fares:

• 1st km: ₹15

• 2nd km: ₹15 + ₹8 = ₹23

• 3rd km: ₹23 + ₹8 = ₹31

• 4th km: ₹31 + ₹8 = ₹39
  and so on...

So the list is: 15, 23, 31, 39, ...

Reason:

Each term increases by ₹8, which is a constant difference.

Answer:
Yes. 15, 23, 31, ... forms an AP as each succeeding term is obtained by adding ₹8 to the preceding term.

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(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Solution:

A vacuum pump removes 1/4 of the air remaining in a cylinder at a time.

Let the initial volume of air = V

Each time, 1/4 of the remaining air is removed, so 3/4 remains.

Volumes after each operation:

• Initial: V

• After 1st pump: (3/4) × V = 3V/4

• After 2nd pump: (3/4) × (3V/4) = (3/4)² × V = 9V/16

• After 3rd pump: (3/4) × (9V/16) = (3/4)³ × V = 27V/64
  and so on...

Pattern:

V, 3V/4, (3/4)² × V, (3/4)³ × V, ...

This is a geometric progression, not an arithmetic one.

Answer:
No. Volumes are V, 3V/4, (3/4)² × V, ... which form a geometric progression, not an arithmetic progression.

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Q1

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

Solution:

The cost of digging a well increases after every metre:

• First metre: ₹150

• Then it increases by ₹50 for each additional metre.

List the costs:

• 1st metre: ₹150

• 2nd metre: ₹150 + ₹50 = ₹200

• 3rd metre: ₹200 + ₹50 = ₹250

• 4th metre: ₹250 + ₹50 = ₹300
  and so on...

So the list is: 150, 200, 250, 300, ...

Reason:

Each term increases by ₹50, a constant difference, so it forms an arithmetic progression.

Answer:
Yes. 150, 200, 250, ... form an AP.

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Q1

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

Solution:

₹10000 is deposited at 8% compound interest per annum.

Formula for compound interest amount:

Amount after n years = 10000 × (1 + 8/100)ⁿ

List of amounts:

• After 1 year: 10000 × (1 + 8/100) = 10000 × 1.08

• After 2 years: 10000 × (1.08)²

• After 3 years: 10000 × (1.08)³

• and so on...

So the list is:
10000 × 1.08, 10000 × (1.08)², 10000 × (1.08)³, ...

This is a geometric progression, not an arithmetic one.

Answer:
No. Amounts are 10000 × 1.08, 10000 × (1.08)², 10000 × (1.08)³, ... which form a geometric progression, not an arithmetic progression.

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Question 2

Write first four terms of the AP, when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

Solution:

• First term a = 10

• Common difference d = 10

Formula for terms of an AP:

nth term = a + (n – 1) × d

First four terms:

• 1st term: 10

• 2nd term: 10 + 10 = 20

• 3rd term: 10 + 20 = 30

• 4th term: 10 + 30 = 40

Answer:
10, 20, 30, 40

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Q2

(ii) a = –2, d = 0

Solution:

• First term a = –2

• Common difference d = 0

First four terms of the AP:

• 1st term: –2

• 2nd term: –2 + 0 = –2

• 3rd term: –2 + 0 = –2

• 4th term: –2 + 0 = –2

Answer:
–2, –2, –2, –2

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Q2

(iii) a = 4, d = – 3

Solution:

• First term a = 4

• Common difference d = –3

First four terms of the AP:

• 1st term: 4

• 2nd term: 4 + (–3) = 1

• 3rd term: 1 + (–3) = –2

• 4th term: –2 + (–3) = –5

Answer:
4, 1, –2, –5

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Q2

(iv) a = – 1, d = ½

Solution:

• First term a = –1

• Common difference d = 1/2

First four terms of the AP:

• 1st term: –1

• 2nd term: –1 + 1/2 = –1/2

• 3rd term: –1/2 + 1/2 = 0

• 4th term: 0 + 1/2 = 1/2

Answer:
–1, –1/2, 0, 1/2

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Q2

(v) a = – 1.25, d = – 0.25

Solution:

• First term a = –1.25

• Common difference d = –0.25

First four terms of the AP:

• 1st term: –1.25

• 2nd term: –1.25 + (–0.25) = –1.50

• 3rd term: –1.50 + (–0.25) = –1.75

• 4th term: –1.75 + (–0.25) = –2.00

Answer:
–1.25, –1.50, –1.75, –2.00

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Q3

For the following APs, write the first term and the common difference:

(i) 3, 1, – 1, – 3, . . .

Solution:

Given AP: 3, 1, –1, –3, ...

First term (a):

a = 3

Common difference (d):

d = 1 – 3 = –2

Answer:
a = 3, d = –2

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Q3

(ii) – 5, – 1, 3, 7, . . .

Solution:

Given AP: –5, –1, 3, 7, ...

First term (a):

a = –5

Common difference (d):

d = –1 – (–5) = –1 + 5 = 4

Answer:
a = –5, d = 4

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Q3

(iii) 1/3, 5/3, 9/3, 13/3, ……

Solution:

Given AP:

1/3, 5/3, 9/3, 13/3, ...

First term (a):

a = 1/3

Common difference (d):

d = 5/3 – 1/3 = 4/3

Answer:
a = 1/3, d = 4/3

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Q3

(iv) 0.6, 1.7, 2.8, 3.9, . . .

Solution:

Given AP:

0.6, 1.7, 2.8, 3.9, ...

First term (a):

a = 0.6

Common difference (d):

d = 1.7 – 0.6 = 1.1

Answer:
a = 0.6, d = 1.1

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Question 4

(i) 2, 4, 8, 16, . . .

Solution:

Given sequence:

2, 4, 8, 16, ...

Check if it forms an AP:

• 4 – 2 = 2

• 8 – 4 = 4

• 16 – 8 = 8

The differences are not equal, so it is not an arithmetic progression.

Answer:
No. The given sequence is not an AP because the differences between consecutive terms are not equal.

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Q4

(ii) 2, 5/2, 3, 7/2, ...

Solution:

Given sequence:

2, 5/2, 3, 7/2, ...

Check if it forms an AP:

• 5/2 – 2 = 1/2

• 3 – 5/2 = 1/2

• 7/2 – 3 = 1/2

Since the common difference is the same, it is an AP.

First term (a):

a = 2

Common difference (d):

d = 1/2

Next three terms:

• 7/2 + 1/2 = 4

• 4 + 1/2 = 9/2

• 9/2 + 1/2 = 5

Answer:
Yes, it is an AP.
Common difference d = 1/2
Next three terms: 4, 9/2, 5

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Q4

(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .

Solution:

Given sequence:

–1.2, –3.2, –5.2, –7.2, ...

Check common difference (d):

–3.2 – (–1.2) = –2
–5.2 – (–3.2) = –2
–7.2 – (–5.2) = –2

So, the common difference is –2.

Next three terms:

–7.2 – 2 = –9.2
–9.2 – 2 = –11.2
–11.2 – 2 = –13.2

Answer:
Yes. d = –2
Next three terms: –9.2, –11.2, –13.2

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Q4

(iv) – 10, – 6, – 2, 2, . . .

Solution:

Given sequence:

–10, –6, –2, 2, ...

Common difference (d):

–6 – (–10) = 4
–2 – (–6) = 4
2 – (–2) = 4

So, d = 4

Next three terms:

2 + 4 = 6
6 + 4 = 10
10 + 4 = 14

Answer:
Yes. d = 4
Next three terms: 6, 10, 14

Q4

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …

Solution:

This is an arithmetic progression (AP) where:

• First term a = 3

• Common difference d = (3 + √2) – 3 = √2

Yes, it is an AP with common difference d = √2.

The next few terms after 3 + 3√2 are:

• 3 + 4√2

• 3 + 5√2

• 3 + 6√2

Answer: Yes. d = √2; 3 + 4√2, 3 + 5√2, 3 + 6√2

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Q4 (vi) 0.2, 0.22, 0.222, 0.2222, …

Solution:

Let's check the difference between terms:

• 0.22 – 0.2 = 0.02

• 0.222 – 0.22 = 0.002

• 0.2222 – 0.222 = 0.0002

The difference is not constant, so this is not an arithmetic progression.

Answer: No.

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Q4 (vii) 0, –4, –8, –12, …

Solution:

Check the common difference:

• (–4) – 0 = –4

• (–8) – (–4) = –4

• (–12) – (–8) = –4

So, it is an arithmetic progression (AP) with:

• First term a = 0

• Common difference d = –4

Next terms:

• –12 + (–4) = –16

• –16 + (–4) = –20

• –20 + (–4) = –24

Answer: Yes. d = –4; –16, –20, –24

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Q4 (viii) –1/2, –1/2, –1/2, –1/2, …

Solution:

All terms are the same, so the common difference is:

• (–1/2) – (–1/2) = 0

This is an arithmetic progression (AP) with:

• Common difference d = 0

Next terms:

• –1/2, –1/2, –1/2

Answer: Yes, d = 0; –1/2, –1/2, –1/2

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Q4 (ix) 1, 3, 9, 27, …

Solution:

Check the differences between terms:

• 3 – 1 = 2

• 9 – 3 = 6

• 27 – 9 = 18

The difference is not constant, so it is not an arithmetic progression (AP).

Answer: No.

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Q4 (x) a, 2a, 3a, 4a, …

Solution:

Check the common difference:

• 2a – a = a

• 3a – 2a = a

• 4a – 3a = a

The difference is constant, so it is an arithmetic progression (AP) with:

• Common difference d = a

Next terms:

• 4a + a = 5a

• 5a + a = 6a

• 6a + a = 7a

Answer: Yes. d = a; 5a, 6a, 7a

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Q4 (xi) a, a², a³, a⁴, …

Solution:

This is a geometric sequence, not arithmetic.

Check the differences:

• a² – a = not equal to a³ – a², etc.

The difference is not constant, so this is not an arithmetic progression (AP).

Answer: No.

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Q4 (xii) √2, √8, √18, √32, …

Solution:

Let's express all terms as multiples of √2:

• √2 = √2

• √8 = √(4×2) = 2√2

• √18 = √(9×2) = 3√2

• √32 = √(16×2) = 4√2

So the sequence becomes:

√2, 2√2, 3√2, 4√2, …

This is an arithmetic progression (AP) with:

• First term = √2

• Common difference d = √2

Next terms:

• 4√2 + √2 = 5√2 = √(25×2) = √50

• 6√2 = √72

• 7√2 = √98

Answer: Yes. d = √2; √50, √72, √98

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Q4 (xiii) √3, √6, √9, √12, …

Solution:

Let's simplify:

• √3 = √3

• √6 = √6

• √9 = 3

• √12 = 2√3

Now check the differences:

• √6 – √3 ≠ 3 – √6

• 3 – √6 ≠ 2√3 – 3

The difference is not constant, so this is not an arithmetic progression (AP).

Answer: No.

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Q4 (xiv) 1², 3², 5², 7², …

Solution:

Step 1: Find the actual numbers

• 1² = 1

• 3² = 9

• 5² = 25

• 7² = 49

So, the sequence becomes:

1, 9, 25, 49, …

Step 2: Check if it is an Arithmetic Progression (AP)

To be an AP, the difference between consecutive terms must be the same.

Now calculate the differences:

• 9 – 1 = 8

• 25 – 9 = 16

• 49 – 25 = 24

Since the differences are not equal, this is not an arithmetic progression.

Answer: No, this is not an AP.

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Q4 (xv) 12, 52, 72, 73…..

Solution:

Step 1: Convert squares to numbers:

• 1² = 1

• 5² = 25

• 7² = 49

• Then comes 73 (already given)

So, the sequence is:

1, 25, 49, 73, …

Step 2: Check the differences between terms:

• 25 – 1 = 24

• 49 – 25 = 24

• 73 – 49 = 24

The common difference is 24, so this is an Arithmetic Progression (AP).

Step 3: Find the next three terms:

• 73 + 24 = 97

• 97 + 24 = 121

• 121 + 24 = 145

Answer: Yes. d = 24; 97, 121, 145

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Exercise 5.2

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

(Refer book for table)

Formula used: nth term of an AP: an = a + (n - 1) × d

(i)

Given:
a = 7, d = 3, n = 8
Find an:
an = 7 + (8 - 1) × 3
an = 7 + 21 = 28

(ii)

Given:
a = -18, n = 10, an = 0
Find d:
0 = -18 + (10 - 1) × d
0 = -18 + 9d
18 = 9d
d = 2

(iii)

Given:
d = -3, n = 18, an = -5
Find a:
-5 = a + (18 - 1) × (-3)
-5 = a - 51
a = -5 + 51 = 46

(iv)

Given:
a = -18.9, d = 2.5, an = 3.6
Find n:
3.6 = -18.9 + (n - 1) × 2.5
3.6 + 18.9 = (n - 1) × 2.5
22.5 = (n - 1) × 2.5
n - 1 = 9
n = 10

(v)

Given:
a = 3.5, d = 0, n = 105
Find an:
an = 3.5 + (105 - 1) × 0
an = 3.5 + 0 = 3.5

Final Answers:

(i) an = 28; (ii) d = 2; (iii) a = 46; (iv) n = 10; (v) an = 3.5

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Q2 Choose the correct choice in the following and justify :

(i) 30th term of the AP: 10, 7, 4, . . . , is

Solution:

To find the 30th term, we use the formula:
an = a + (n - 1) × d

Where:

• a = first term = 10

• d = common difference = 7 - 10 = -3

• n = 30

Now, plug in the values:
a30 = 10 + (30 - 1) × (-3)
a30 = 10 + 29 × (-3)
a30 = 10 - 87
a30 = -77

Answer: The 30th term is -77 (option C)

Justification:
Since the AP has a common difference of -3, each term decreases by 3. So, the 30th term is -77 using the AP formula.

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Q2 (ii) 11th term of the AP: – 3, -1/2, 2, . . ., is

Solution:

Step 1: Identify the first term and common difference

• First term (a) = –3

• Common difference (d) = –1/2 – (–3) = –1/2 + 3 = 5/2

Step 2: Use the formula to find the 11th term:
an = a + (n – 1) × d

a₁₁ = –3 + (11 – 1) × (5/2)
a₁₁ = –3 + 10 × (5/2)
a₁₁ = –3 + 25
a₁₁ = 22

Answer: The 11th term is 22. (option B)

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Q3 In the following APs, find the missing terms in the blanks:

(i) 2, ___, 26

Let’s call the missing term x.

Since it’s an AP, the difference between consecutive terms must be the same.
So:

x – 2 = 26 – x

Now solve:

x – 2 = 26 – x
x + x = 26 + 2
2x = 28
x = 14

Answer: The missing term is 14.
So, the AP is: 2, 14, 26.

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Q3 (ii) ___, 13, ___, 3

Let’s call the first term a, and the common difference d.

That means the sequence looks like this:

• 1st term: a

• 2nd term: a + d = 13

• 3rd term: a + 2d

• 4th term: a + 3d = 3

Step 1: Use the known terms to form equations

From 2nd term:
a + d = 13 → (Equation 1)

From 4th term:
a + 3d = 3 → (Equation 2)

Step 2: Subtract Equation 1 from Equation 2

(a + 3d) – (a + d) = 3 – 13
a + 3d – a – d = –10
2d = –10
d = –5

Step 3: Find a

From Equation 1:
a + d = 13
a – 5 = 13
a = 18

Step 4: Fill in the AP

• 1st term: a = 18

• 2nd term: 18 + (–5) = 13

• 3rd term: 13 + (–5) = 8

• 4th term: 8 + (–5) = 3

Final Answer: 18, 13, 8, 3

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Q3 (iii) 5, ___, ___, 9(1/2)

Solution:
Step 1: Let the number of terms be:

• 1st term: 5

• 2nd term: ?

• 3rd term: ?

• 4th term: 9½ = 19/2

Let the common difference be d.

Step 2: Use the AP formula

The nth term formula is:
an = a + (n – 1) × d

Here,
a = 5
a₄ = 5 + 3d = 19/2

Now solve:

5 + 3d = 19/2
Convert 5 to fraction: 10/2 + 3d = 19/2
3d = 19/2 – 10/2 = 9/2
d = (9/2) ÷ 3 = 3/2

Step 3: Find the missing terms

• 2nd term = 5 + 3/2 = 13/2

• 3rd term = 13/2 + 3/2 = 16/2 = 8

Answer: 5, 13/2, 8, 9½ or 5, 6½, 8, 9½

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Q3 (iv) –4, ___, ___, ___, ___, 6

Solution:

Step 1: Use the AP formula

an = a + (n – 1) × d

Here,

• First term (a) = –4

• Sixth term (a₆) = 6

• Number of terms (n) = 6

So:
a₆ = a + 5d
6 = –4 + 5d
5d = 10
d = 2

Step 2: Find missing terms using d = 2

• 1st term: –4

• 2nd term: –4 + 2 = –2

• 3rd term: –2 + 2 = 0

• 4th term: 0 + 2 = 2

• 5th term: 2 + 2 = 4

• 6th term: 4 + 2 = 6

Answer: –4, –2, 0, 2, 4, 6

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Q3 (v) ___, 38, ___, ___, ___, –22

Solution:

This is a sequence of 6 terms:
Let the terms be:
a₁, a₂, a₃, a₄, a₅, a₆

From the question:

• a₂ = 38

• a₆ = –22

We need to find a₁, a₃, a₄, a₅

Step 1: Use the AP formula:

General term of an AP:
aₙ = a + (n – 1)d

Let’s assume first term = a, common difference = d

From a₂ = 38:
→ a + (2 – 1)d = 38
→ a + d = 38 → (Equation 1)

From a₆ = –22:
→ a + 5d = –22 → (Equation 2)

Step 2: Solve the equations

From Equation 1:
a = 38 – d

Substitute in Equation 2:

(38 – d) + 5d = –22
→ 38 + 4d = –22
→ 4d = –60
→ d = –15

Now, substitute back to find a:
a = 38 – (–15) = 38 + 15 = 53

Step 3: Find all terms

Now that we know:

• a = 53

• d = –15

Use the formula: aₙ = a + (n – 1)d

• a₁ = 53

• a₂ = 53 + (1)d = 53 + (–15) = 38

• a₃ = 53 + 2(–15) = 53 – 30 = 23

• a₄ = 53 + 3(–15) = 53 – 45 = 8

• a₅ = 53 + 4(–15) = 53 – 60 = –7

• a₆ = 53 + 5(–15) = 53 – 75 = –22

Sequence: 53, 38, 23, 8, –7, –22

Final Answer:

53, 38, 23, 8, –7, –22
Missing numbers are: 53, 23, 8, –7

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Q4 Which term of the AP : 3, 8, 13, 18, . . . ,is 78?

Solution:

Given AP: 3, 8, 13, 18, ...

We are asked to find which term is 78.

Step 1: Identify the first term and common difference
First term (a) = 3
Common difference (d) = 8 - 3 = 5

Step 2: Use the nth term formula of AP
Tn = a + (n - 1) × d

We are given Tn = 78, so:

78 = 3 + (n - 1) × 5

Step 3: Solve the equation
78 - 3 = (n - 1) × 5
75 = (n - 1) × 5
75 ÷ 5 = n - 1
15 = n - 1
n = 16

Answer: 78 is the 16th term of the AP.

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Q5 Find the number of terms in each of the following APs :

(i) 7, 13, 19, ..., 205

Solution:

Given AP: 7, 13, 19, ..., 205

Step 1: Identify the first term and common difference
First term (a) = 7
Common difference (d) = 13 - 7 = 6
Last term (l) = 205

Step 2: Use the nth term formula
Tn = a + (n - 1) × d
Substitute the values:

205 = 7 + (n - 1) × 6

Step 3: Solve the equation
205 - 7 = (n - 1) × 6
198 = (n - 1) × 6
198 ÷ 6 = n - 1
33 = n - 1
n = 34

Answer: The number of terms is 34.

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Q5

(ii) 18, 15½, 13, ..., –47

Solution:

Step 1: Identify the first term and common difference
First term (a) = 18
Second term = 15½
Common difference (d) = 15½ - 18 = –2½
So, d = –2.5
Last term (l) = –47

Step 2: Use the nth term formula
Tn = a + (n - 1) × d
Substitute the values:

–47 = 18 + (n - 1) × (–2.5)

Step 3: Solve the equation
–47 - 18 = (n - 1) × (–2.5)
–65 = (n - 1) × (–2.5)
Divide both sides by –2.5:

–65 ÷ –2.5 = n - 1
26 = n - 1
n = 27

Answer: The number of terms is 27.

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Q6 Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .

Solution:

Given AP: 11, 8, 5, 2, ...

Step 1: Identify the first term and common difference
First term (a) = 11
Common difference (d) = 8 – 11 = –3
We are checking if –150 is any term (Tn) in the AP.

Step 2: Use the nth term formula
Tn = a + (n - 1) × d
Substitute the known values:

–150 = 11 + (n - 1) × (–3)

Step 3: Solve the equation
–150 – 11 = (n - 1) × (–3)
–161 = (n - 1) × (–3)
Divide both sides by –3:

–161 ÷ –3 = n - 1
53.67 = n - 1
n = 54.67

Step 4: Conclusion
Since n is not a whole number, –150 is not a term of the AP.

Answer: No, –150 is not a term of the AP.

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Q7 Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:

Given:
11th term (T₁₁) = 38
16th term (T₁₆) = 73
We are asked to find the 31st term (T₃₁).

Step 1: Use the nth term formula
Tn = a + (n - 1) × d

From T₁₁ = a + 10d = 38 → (Equation 1)
From T₁₆ = a + 15d = 73 → (Equation 2)

Step 2: Subtract Equation 1 from Equation 2
(a + 15d) – (a + 10d) = 73 – 38
a + 15d – a – 10d = 35
5d = 35
d = 7

Step 3: Substitute d into Equation 1 to find a
a + 10 × 7 = 38
a + 70 = 38
a = 38 – 70
a = –32

Step 4: Find the 31st term
T₃₁ = a + (31 - 1) × d
T₃₁ = –32 + 30 × 7
T₃₁ = –32 + 210
T₃₁ = 178

Answer: The 31st term is 178.

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Q8 An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Solution:

Given:
Total number of terms = 50
3rd term (T₃) = 12
Last term (T₅₀) = 106
We are asked to find the 29th term (T₂₉).

Step 1: Use the nth term formula
Tn = a + (n - 1) × d

We know:
T₃ = a + 2d = 12 → (Equation 1)
T₅₀ = a + 49d = 106 → (Equation 2)

Step 2: Subtract Equation 1 from Equation 2
(a + 49d) – (a + 2d) = 106 – 12
a + 49d – a – 2d = 94
47d = 94
d = 2

Step 3: Substitute d into Equation 1 to find a
a + 2 × 2 = 12
a + 4 = 12
a = 8

Step 4: Find the 29th term
T₂₉ = a + (29 – 1) × d
T₂₉ = 8 + 28 × 2
T₂₉ = 8 + 56
T₂₉ = 64

Answer: The 29th term is 64.

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Q9 If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?

Solution:

Given:
3rd term (T₃) = 4
9th term (T₉) = –8

We are asked to find which term is zero (Tn = 0).

Step 1: Use the nth term formula
Tn = a + (n - 1) × d

From T₃ = a + 2d = 4 → (Equation 1)
From T₉ = a + 8d = –8 → (Equation 2)

Step 2: Subtract Equation 1 from Equation 2
(a + 8d) – (a + 2d) = –8 – 4
a + 8d – a – 2d = –12
6d = –12
d = –2

Step 3: Substitute d into Equation 1 to find a
a + 2 × (–2) = 4
a – 4 = 4
a = 8

Step 4: Find the term number for which Tn = 0
Tn = a + (n – 1) × d
0 = 8 + (n – 1) × (–2)

Solve:
0 = 8 – 2(n – 1)
0 = 8 – 2n + 2
0 = 10 – 2n
2n = 10
n = 5

Answer: Zero is the 5th term of the AP.

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Q10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Solution:

Given:
The 17th term exceeds the 10th term by 7.
We are asked to find the common difference (d).

Step 1: Use the nth term formula
Tn = a + (n - 1) × d

T₁₇ = a + 16d
T₁₀ = a + 9d

According to the question:
T₁₇ – T₁₀ = 7

Substitute the expressions:
(a + 16d) – (a + 9d) = 7
a + 16d – a – 9d = 7
7d = 7
d = 1

Answer: The common difference is 1.

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Q11

Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?

Solution:

Given AP: 3, 15, 27, 39, ...
We are asked: Which term is 132 more than the 54th term?

Step 1: Identify the first term and common difference
First term (a) = 3
Common difference (d) = 15 – 3 = 12

Step 2: Use the nth term formula
Tn = a + (n – 1) × d

Find the 54th term:
T₅₄ = 3 + (54 – 1) × 12
T₅₄ = 3 + 53 × 12
T₅₄ = 3 + 636 = 639

Now, let Tn be the term which is 132 more than T₅₄:
Tn = 639 + 132 = 771

Step 3: Use the nth term formula again to find n
771 = 3 + (n – 1) × 12
771 – 3 = (n – 1) × 12
768 = (n – 1) × 12
n – 1 = 768 ÷ 12 = 64
n = 65

Answer: The required term is the 65th term.

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Q12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution:

Given:
Two APs have the same common difference.
The difference between their 100th terms is 100.
We are asked to find the difference between their 1000th terms.

Step 1: Use the nth term formula
Tn = a + (n – 1) × d

Let:

• First AP: Tn = a₁ + (n – 1) × d

• Second AP: Tn = a₂ + (n – 1) × d

Both APs have the same common difference d.

Step 2: Use the given condition
Difference between 100th terms:
T₁₀₀ (first) – T₁₀₀ (second) = 100

So:
[a₁ + 99d] – [a₂ + 99d] = 100
a₁ – a₂ = 100

Step 3: Find the difference between 1000th terms
T₁₀₀₀ (first) – T₁₀₀₀ (second)
= [a₁ + 999d] – [a₂ + 999d]
= a₁ – a₂
= 100

Answer: The difference between the 1000th terms is 100.

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Q13

How many three-digit numbers are divisible by 7?

Solution:

Step 1: Find the smallest and largest three-digit numbers

• Smallest three-digit number = 100

• Largest three-digit number = 999

Step 2: Find the first three-digit number divisible by 7
Divide 100 by 7:
100 ÷ 7 ≈ 14.28 → next whole number is 15
First number = 15 × 7 = 105

Step 3: Find the last three-digit number divisible by 7
Divide 999 by 7:
999 ÷ 7 ≈ 142.71 → take whole number part = 142
Last number = 142 × 7 = 994

Step 4: Use the AP formula to find number of terms
The numbers divisible by 7 form an AP:
105, 112, 119, ..., 994

This is an AP with:

• First term (a) = 105

• Last term (l) = 994

• Common difference (d) = 7

Use the formula:
n = ((l – a) ÷ d) + 1
n = ((994 – 105) ÷ 7) + 1
n = (889 ÷ 7) + 1
n = 127 + 1
n = 128

Answer: There are 128 three-digit numbers divisible by 7.

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Q14

How many multiples of 4 lie between 10 and 250?

Solution:

Step 1: Find the first multiple of 4 greater than 10
10 ÷ 4 = 2.5 → next whole number = 3
First multiple = 3 × 4 = 12

Step 2: Find the last multiple of 4 less than 250
250 ÷ 4 = 62.5 → take whole number part = 62
Last multiple = 62 × 4 = 248

Step 3: Use the AP formula to find number of terms
Multiples of 4 between 10 and 250:
12, 16, 20, ..., 248

This is an AP with:

• First term (a) = 12

• Last term (l) = 248

• Common difference (d) = 4

Use the formula:
n = ((l – a) ÷ d) + 1
n = ((248 – 12) ÷ 4) + 1
n = (236 ÷ 4) + 1
n = 59 + 1
n = 60

Answer: There are 60 multiples of 4 between 10 and 250.

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Q15

For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?

Solution:

Given APs:

• First AP: 63, 65, 67, ...

• Second AP: 3, 10, 17, ...

Step 1: Identify the first term and common difference of each AP
First AP:

• a₁ = 63

• d₁ = 65 – 63 = 2

Second AP:

• a₂ = 3

• d₂ = 10 – 3 = 7

Step 2: Write nth term for both APs

Tn (first AP) = 63 + (n – 1) × 2
Tn (second AP) = 3 + (n – 1) × 7

Set both nth terms equal:

63 + (n – 1) × 2 = 3 + (n – 1) × 7

Step 3: Solve the equation

Expand both sides:

63 + 2n – 2 = 3 + 7n – 7
61 + 2n = –4 + 7n

Now solve:

61 + 2n = –4 + 7n
61 + 4 = 7n – 2n
65 = 5n
n = 13

Answer: The nth terms are equal when n = 13.

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Q16

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution:

Step 1: Use the general term formula
Tn = a + (n – 1) × d

Given:
T₃ = a + 2d = 16 → (Equation 1)
T₇ – T₅ = 12
→ (a + 6d) – (a + 4d) = 12
a + 6d – a – 4d = 12
2d = 12
d = 6

Step 2: Substitute d into Equation 1 to find a
a + 2 × 6 = 16
a + 12 = 16
a = 4

Step 3: Write the AP
Now that we have a = 4 and d = 6:
AP = 4, 10, 16, 22, ...

Answer: The required AP is 4, 10, 16, 22, ...

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Q17

Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.

Solution:

Step 1: Identify the first term, last term, and common difference
First term (a) = 3
Common difference (d) = 8 – 3 = 5
Last term (l) = 253

Step 2: Find the total number of terms (n)
Use the nth term formula:
Tn = a + (n – 1) × d
253 = 3 + (n – 1) × 5
253 – 3 = (n – 1) × 5
250 = (n – 1) × 5
n – 1 = 50
n = 51

So, the AP has 51 terms.

Step 3: Find the 20th term from the last
The 20th term from the last is the (51 – 20 + 1) = 32nd term from the beginning.

Use the formula:
T₃₂ = a + (32 – 1) × d
T₃₂ = 3 + 31 × 5
T₃₂ = 3 + 155 = 158

Answer: The 20th term from the last is 158.

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Q18

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution:

Step 1: Use the nth term formula
Tn = a + (n – 1) × d

Given:
T₄ + T₈ = 24
T₆ + T₁₀ = 44

Step 2: Write expressions using the formula
T₄ = a + 3d
T₈ = a + 7d
So:
(a + 3d) + (a + 7d) = 24
2a + 10d = 24 → (Equation 1)

T₆ = a + 5d
T₁₀ = a + 9d
So:
(a + 5d) + (a + 9d) = 44
2a + 14d = 44 → (Equation 2)

Step 3: Subtract the equations
Equation 2 – Equation 1:
(2a + 14d) – (2a + 10d) = 44 – 24
2a – 2a + 4d = 20
4d = 20
d = 5

Step 4: Substitute d into Equation 1 to find a
2a + 10 × 5 = 24
2a + 50 = 24
2a = 24 – 50 = –26
a = –13

Step 5: Write the first three terms
a = –13, d = 5
First term = –13
Second term = –13 + 5 = –8
Third term = –8 + 5 = –3

Answer: The first three terms are –13, –8, –3

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Q19

Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

Solution:

Step 1: Identify the AP terms
This is an Arithmetic Progression (AP) where:

• First term (a) = ₹5000

• Common difference (d) = ₹200

• Final salary = ₹7000

Let his salary in the n-th year be ₹7000.

Step 2: Use the AP formula
Tn = a + (n – 1) × d
7000 = 5000 + (n – 1) × 200

Step 3: Solve the equation
7000 – 5000 = (n – 1) × 200
2000 = (n – 1) × 200
(n – 1) = 2000 ÷ 200 = 10
n = 11

Step 4: Find the year
He started in 1995, so the 11th year = 1995 + 10 = 2005

Answer: His income reached ₹7000 in the 11th year, i.e., in 2005.

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Q20

Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.

Solution:

Step 1: Recognize the AP
This is an Arithmetic Progression where:

• First term (a) = ₹5

• Common difference (d) = ₹1.75

• nth term (Tn) = ₹20.75

Step 2: Use the AP formula
Tn = a + (n – 1) × d
20.75 = 5 + (n – 1) × 1.75

Step 3: Solve the equation
20.75 – 5 = (n – 1) × 1.75
15.75 = (n – 1) × 1.75
(n – 1) = 15.75 ÷ 1.75 = 9
n = 9 + 1 = 10

Answer: The value of n is 10.

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Exercise 5.3

Q1.

Find the sum of the following APs:

(i) 2, 7, 12, . . ., to 10 terms.

Solution:

Step 1: Identify the AP details

• First term (a) = 2

• Common difference (d) = 7 – 2 = 5

• Number of terms (n) = 10

Step 2: Use the sum formula for AP
Sn = n/2 × [2a + (n – 1) × d]

Substitute the values:
S₁₀ = 10/2 × [2 × 2 + (10 – 1) × 5]
= 5 × [4 + 9 × 5]
= 5 × [4 + 45]
= 5 × 49
= 245

Answer: The sum of the AP is 245

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Q1.

(ii) –37, –33, –29, . . ., to 12 terms.

Solution:

Step 1: Identify the AP details

• First term (a) = –37

• Common difference (d) = –33 – (–37) = 4

• Number of terms (n) = 12

Step 2: Use the sum formula for AP
Sn = n/2 × [2a + (n – 1) × d]

Substitute the values:
S₁₂ = 12/2 × [2 × (–37) + (12 – 1) × 4]
= 6 × [–74 + 11 × 4]
= 6 × [–74 + 44]
= 6 × (–30)
= –180

Answer: The sum of the AP is –180

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Q1

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms

Solution:

Step 1: Identify first term and common difference

• First term (a) = 0.6

• Common difference (d) = 1.7 - 0.6 = 1.1

• Number of terms (n) = 100

Step 2: Use the formula for sum of n terms

Sum = (n / 2) × [2a + (n - 1)d]

Substitute the values:
= (100 / 2) × [2 × 0.6 + (100 - 1) × 1.1]
= 50 × [1.2 + 99 × 1.1]
= 50 × [1.2 + 108.9]
= 50 × 110.1
= 5505

Answer: 5505

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Q1

(iv) 1/15, 1/12, 1/10, ..., to 11 terms

Solution:

Step 1: Identify first term and common difference

• First term (a) = 1/15

• Second term = 1/12

• Common difference (d) = 1/12 – 1/15
  = (5 – 4)/60 = 1/60

• Number of terms (n) = 11

Step 2: Use the formula for sum of n terms

Sum = (n / 2) × [2a + (n - 1)d]

Substitute the values:
= (11 / 2) × [2 × (1/15) + (11 – 1) × (1/60)]
= (11 / 2) × [(2/15) + (10/60)]
= (11 / 2) × [(2/15) + (1/6)]

Now make common denominator:
= (11 / 2) × [(4/30) + (5/30)]
= (11 / 2) × (9/30)
= (11 / 2) × (3/10)
= (33 / 20)

Answer: 33/20

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Q2

Find the sums given below :

(i) 7 + 10½ + 14 + … + 84

Solution:

Step 1: Convert 10½ to improper fraction or decimal

10½ = 21/2 = 10.5

So the A.P. becomes:
7, 10.5, 14, … , 84

Step 2: Identify first term and common difference

• First term (a) = 7

• Common difference (d) = 10.5 – 7 = 3.5

• Last term (l) = 84

Step 3: Find the number of terms (n)

Use the formula for the nth term:
l = a + (n – 1) × d

84 = 7 + (n – 1) × 3.5
=> 84 – 7 = (n – 1) × 3.5
=> 77 = (n – 1) × 3.5
=> (n – 1) = 77 ÷ 3.5 = 22
=> n = 23

Step 4: Use the formula for sum of n terms

Sum = (n / 2) × (a + l)
= (23 / 2) × (7 + 84)
= (23 / 2) × 91
= (23 × 91) / 2
= 2093 / 2
= 1046½

Answer: 1046½

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Q2

(ii) 34 + 32 + 30 + ... + 10

Solution:

Step 1: Identify first term and common difference

• First term (a) = 34

• Common difference (d) = 32 – 34 = –2

• Last term (l) = 10

Step 2: Find number of terms (n)

Use the formula for the nth term:
l = a + (n – 1) × d

10 = 34 + (n – 1) × (–2)
=> 10 – 34 = (n – 1) × (–2)
=> –24 = (n – 1) × (–2)
=> n – 1 = 12
=> n = 13

Step 3: Use sum formula

Sum = (n / 2) × (a + l)
= (13 / 2) × (34 + 10)
= (13 / 2) × 44
= (13 × 44) / 2 = 572 / 2
= 286

Answer: 286

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Q2

(iii) –5 + (–8) + (–11) + . . . + (–230)

Solution:

Step 1: Identify first term and common difference

• First term (a) = –5

• Common difference (d) = –8 – (–5) = –3

• Last term (l) = –230

Step 2: Find the number of terms (n)

Use the nth term formula:
l = a + (n – 1) × d

–230 = –5 + (n – 1) × (–3)
=> –230 + 5 = (n – 1) × (–3)
=> –225 = (n – 1) × (–3)
=> n – 1 = –225 ÷ (–3) = 75
=> n = 76

Step 3: Use the sum formula

Sum = (n / 2) × (a + l)
= (76 / 2) × (–5 + (–230))
= 38 × (–235)
= –8930

Answer: –8930

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Q3

(i) given a = 5, d = 3, an = 50, find n and Sn .

Solution:

We are given:

• First term (a) = 5

• Common difference (d) = 3

• nth term (aₙ) = 50

We need to find:

• Number of terms (n)

• Sum of n terms (Sₙ)

Step 1: Use nth term formula

aₙ = a + (n – 1) × d
50 = 5 + (n – 1) × 3
=> 50 – 5 = (n – 1) × 3
=> 45 = (n – 1) × 3
=> n – 1 = 15
=> n = 16

Step 2: Use sum formula

Sₙ = (n / 2) × (a + aₙ)
= (16 / 2) × (5 + 50)
= 8 × 55 = 440

Final Answer: n = 16, Sₙ = 440

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Q3

(ii) given a = 7, a13 = 35, find d and S13

Solution:

We are given:

• First term (a) = 7

• 13th term (a₁₃) = 35
  We need to find:

• Common difference (d)

• Sum of first 13 terms (S₁₃)

Step 1: Use nth term formula to find d

aₙ = a + (n – 1) × d
35 = 7 + (13 – 1) × d
=> 35 = 7 + 12d
=> 35 – 7 = 12d
=> 28 = 12d
=> d = 28 ÷ 12 = 7/3

Step 2: Use sum formula to find S₁₃

Sₙ = (n / 2) × [2a + (n – 1)d]
S₁₃ = (13 / 2) × [2 × 7 + 12 × (7/3)]
= (13 / 2) × [14 + 84/3]
= (13 / 2) × [14 + 28]
= (13 / 2) × 42
= 13 × 21 = 273

Final Answer: d = 7/3, S₁₃ = 273

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Q3

(iii)

given a12 = 37, d = 3, find a and S12

Solution:

We are given:

• 12th term (a₁₂) = 37

• Common difference (d) = 3

We need to find:

• First term (a)

• Sum of first 12 terms (S₁₂)

Step 1: Use nth term formula to find a

aₙ = a + (n – 1) × d

37 = a + (12 – 1) × 3
=> 37 = a + 11 × 3
=> 37 = a + 33
=> a = 37 – 33 = 4

Step 2: Use sum formula to find S₁₂

Sₙ = (n / 2) × [2a + (n – 1)d]

S₁₂ = (12 / 2) × [2 × 4 + 11 × 3]
= 6 × [8 + 33]
= 6 × 41 = 246

Final Answer: a = 4, S₁₂ = 246

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Q3

(iv)

given a3 = 15, S10 = 125, find d and a10.

Solution:

Given:
a₃ = 15, S₁₀ = 125

Find d and a₁₀

Step 1: Use a₃ = a + 2d
So,
a + 2d = 15 → (Equation 1)

Step 2: Use the AP sum formula
Sₙ = (n/2)[2a + (n – 1)d]
Substitute values:
S₁₀ = (10/2)[2a + 9d] = 125
5[2a + 9d] = 125
2a + 9d = 25 → (Equation 2)

Step 3: Solve the two equations

From Equation 1:
a + 2d = 15 → Multiply by 2
2a + 4d = 30 → (Equation 3)

Now subtract Equation 2 from Equation 3:
(2a + 4d) – (2a + 9d) = 30 – 25
–5d = 5 → d = –1

Step 4: Find a

a + 2d = 15
a + 2(–1) = 15
a – 2 = 15 → a = 17

Step 5: Find a₁₀

a₁₀ = a + 9d
a₁₀ = 17 + 9(–1) = 17 – 9 = 8

Answer: d = –1, a₁₀ = 8

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Q3

(v) given d = 5, S9 = 75, find a and a9

Solution:

Given:
d = 5, S₉ = 75

Find a and a₉

Step 1: Use the AP sum formula
Sₙ = (n/2) × [2a + (n – 1)d]

Substitute values:
S₉ = (9/2) × [2a + 8d] = 75
(9/2) × [2a + 40] = 75

Multiply both sides by 2:
9 × (2a + 40) = 150
2a + 40 = 150 ÷ 9 = 50/3
2a = 50/3 – 40 = 50/3 – 120/3 = –70/3
a = –70/6 = –35/3

Step 2: Find a₉
a₉ = a + 8d = –35/3 + 8×5 = –35/3 + 40
Convert 40 to fraction:
–35/3 + 120/3 = 85/3

Answer: a = –35/3, a₉ = 85/3

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Q3

(vi) given a = 2, d = 8, Sn = 90, find n and an

Solution:

Given:
a = 2, d = 8, Sₙ = 90

Find n and aₙ

Step 1: Use the AP sum formula

Sₙ = (n/2) × [2a + (n – 1)d]
Substitute the values:
90 = (n/2) × [2×2 + (n – 1)×8]
90 = (n/2) × [4 + 8n – 8]
90 = (n/2) × (8n – 4)

Now multiply both sides by 2:
180 = n × (8n – 4)
180 = 8n² – 4n

Bring all terms to one side:
8n² – 4n – 180 = 0

Divide the whole equation by 4:
2n² – n – 45 = 0

Step 2: Solve the quadratic equation

2n² – n – 45 = 0
Use factorization:

2n² – 10n + 9n – 45 = 0
(2n² – 10n) + (9n – 45) = 0
2n(n – 5) + 9(n – 5) = 0
(2n + 9)(n – 5) = 0

So, n = 5 or n = –9/2 (reject, n must be positive)

n = 5

Step 3: Find aₙ

aₙ = a + (n – 1)d
a₅ = 2 + (5 – 1)×8 = 2 + 32 = 34

Answer: n = 5, aₙ = 34

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Q3

(vii) given a = 8, an = 62, Sn = 210, find n and d

Solution:

We are given:

• a = 8

• aₙ = 62

• Sₙ = 210

We use the formulas:

1. aₙ = a + (n – 1) × d

2. Sₙ = n/2 × (a + aₙ)

Step 1: Use the sum formula to find n

Sₙ = n/2 × (a + aₙ)
210 = n/2 × (8 + 62)
210 = n/2 × 70
Multiply both sides by 2:
420 = 70n
Divide both sides by 70:
n = 6

Step 2: Use the aₙ formula to find d

aₙ = a + (n – 1) × d
62 = 8 + (6 – 1) × d
62 = 8 + 5d
Subtract 8 from both sides:
54 = 5d
Divide both sides by 5:
d = 54/5

Answer: n = 6, d = 54/5

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Q3

(viii) given an = 4, d = 2, Sn = –14, find n and a.

Solution:

We are given:

• aₙ = 4

• d = 2

• Sₙ = –14

We use the formulas:

1. aₙ = a + (n – 1) × d

2. Sₙ = n/2 × (a + aₙ)

Step 1: Use the sum formula to find n

Sₙ = n/2 × (a + aₙ)
–14 = n/2 × (a + 4)
Multiply both sides by 2:
–28 = n × (a + 4)
→ Equation (1): –28 = n(a + 4)

Step 2: Use the aₙ formula

aₙ = a + (n – 1) × d
4 = a + (n – 1) × 2
→ Equation (2): 4 = a + 2(n – 1)

Simplify:
4 = a + 2n – 2
Add 2 to both sides:
6 = a + 2n
→ Equation (3): a = 6 – 2n

Step 3: Substitute Equation (3) into Equation (1)

–28 = n × (6 – 2n + 4)
–28 = n × (10 – 2n)
–28 = 10n – 2n²
Bring all terms to one side:
2n² – 10n – 28 = 0
Divide entire equation by 2:
n² – 5n – 14 = 0

Factor the quadratic:
(n – 7)(n + 2) = 0
n = 7 or n = –2 (reject –2)

So, n = 7

Step 4: Find a using Equation (3)

a = 6 – 2n
a = 6 – 2×7 = 6 – 14 = –8

Answer: n = 7, a = –8

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Q3

(ix)

given a = 3, n = 8, S = 192, find d

Solution:

We are given:

• a = 3

• n = 8

• Sₙ = 192

We use the sum formula of an arithmetic progression:
Sₙ = n/2 × [2a + (n – 1) × d]

Step 1: Substitute the known values

192 = 8/2 × [2×3 + (8 – 1) × d]
192 = 4 × [6 + 7d]
192 = 4 × (6 + 7d)
192 = 24 + 28d

Step 2: Solve for d

Subtract 24 from both sides:
168 = 28d
Divide both sides by 28:
d = 6

Answer: d = 6

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Q3

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Solution:

We are given:

• l = 28 (last term)

• Sₙ = 144 (sum of n terms)

• n = 9 (number of terms)

We use the formula:
Sₙ = n/2 × (a + l)

Step 1: Substitute the values

144 = 9/2 × (a + 28)
Multiply both sides by 2:
288 = 9 × (a + 28)
Divide both sides by 9:
32 = a + 28

Step 2: Solve for a

a = 32 – 28
a = 4

Answer: a = 4

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Q4

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution:

We are given the arithmetic progression:
9, 17, 25, ...

This means:

• First term, a = 9

• Common difference, d = 17 – 9 = 8

• Sum, Sₙ = 636

We use the sum formula:
Sₙ = n/2 × [2a + (n – 1) × d]

Substitute the values:
636 = n/2 × [2×9 + (n – 1)×8]
636 = n/2 × [18 + 8(n – 1)]
636 = n/2 × [18 + 8n – 8]
636 = n/2 × (8n + 10)

Multiply both sides by 2:
1272 = n × (8n + 10)

Expand: 1272 = 8n² + 10n

Bring all terms to one side: 8n² + 10n – 1272 = 0

Divide whole equation by 2 to simplify: 4n² + 5n – 636 = 0

Now solve the quadratic: 4n² + 5n – 636 = 0

Using the quadratic formula:
n = [–5 ± √(5² + 4×4×636)] / (2×4)
n = [–5 ± √(25 + 10176)] / 8
n = [–5 ± √10201] / 8
√10201 = 101

n = [–5 ± 101] / 8
So, n = (96)/8 = 12 or (–106)/8 (not valid)

Answer: 12 terms must be taken to get a sum of 636.

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Q5

The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

We are given:

• First term a = 5

• Last term l = 45

• Sum of n terms Sₙ = 400

We use the formula:
Sₙ = n/2 × (a + l)

Step 1: Find the number of terms (n)

400 = n/2 × (5 + 45)
400 = n/2 × 50
Multiply both sides by 2:
800 = 50n
Divide both sides by 50:
n = 16

Step 2: Find the common difference (d)
We use the formula for the last term:
l = a + (n – 1) × d

45 = 5 + (16 – 1) × d
45 = 5 + 15d
Subtract 5 from both sides:
40 = 15d
Divide both sides by 15:
d = 8/3

Answer:
Number of terms (n) = 16
Common difference (d) = 8/3

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Q6

The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

We are given:

• First term a = 17

• Last term l = 350

• Common difference d = 9

We use the formula for the last term:
l = a + (n – 1) × d

Step 1: Find the number of terms (n)

350 = 17 + (n – 1) × 9
Subtract 17:
333 = (n – 1) × 9
Divide both sides by 9:
37 = n – 1
Add 1 to both sides:
n = 38

Step 2: Use the sum formula

Sₙ = n/2 × (a + l)
S = 38/2 × (17 + 350)
S = 19 × 367
S = 6973

Answer: n = 38, S = 6973

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Q7

Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution:

We are given:

• Common difference d = 7

• 22nd term (a₂₂) = 149

• Number of terms n = 22

We use the formula for the n-th term:
aₙ = a + (n – 1) × d

Step 1: Find the first term (a)

a₂₂ = a + 21 × 7
149 = a + 147
Subtract 147 from both sides:
a = 2

Step 2: Use the sum formula

Sₙ = n/2 × (a + aₙ)
S₂₂ = 22/2 × (2 + 149)
S₂₂ = 11 × 151
S₂₂ = 1661

Answer: Sum = 1661

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Q8

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively

Solution:

We are given:

• Second term = 14

• Third term = 18

• We need to find the sum of first 51 terms (S₅₁)

Step 1: Find the common difference (d)

Third term – Second term = 18 – 14 = 4
So, d = 4

Step 2: Find the first term (a)

Second term = a + d
14 = a + 4
a = 14 – 4 = 10

Step 3: Use the sum formula

Sₙ = n/2 × [2a + (n – 1) × d]

S₅₁ = 51/2 × [2×10 + (51 – 1)×4]
S₅₁ = 51/2 × [20 + 50×4]
S₅₁ = 51/2 × [20 + 200]
S₅₁ = 51/2 × 220
S₅₁ = (51 × 220) / 2 = 11220 / 2 = 5610

Answer: S₅₁ = 5610

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Q9

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution:

We are given:

• Sum of first 7 terms (S₇) = 49

• Sum of first 17 terms (S₁₇) = 289

• We are to find a general formula for the sum of the first n terms: Sₙ = ?

Step 1: Use the sum formula

Sₙ = n/2 × [2a + (n – 1)d]

So, for S₇:
49 = 7/2 × [2a + 6d] → Multiply both sides by 2:
98 = 7 × (2a + 6d)

Divide both sides by 7:
2a + 6d = 14 → Equation (1)

Now for S₁₇:
289 = 17/2 × [2a + 16d] → Multiply both sides by 2:
578 = 17 × (2a + 16d)
Divide both sides by 17:
2a + 16d = 34 → Equation (2)

Step 2: Subtract Equation (1) from Equation (2)

(2a + 16d) – (2a + 6d) = 34 – 14
10d = 20
d = 2

Substitute into Equation (1):
2a + 6×2 = 14
2a + 12 = 14
2a = 2
a = 1

Step 3: Write the sum formula

Sₙ = n/2 × [2a + (n – 1)d]
Substitute a = 1, d = 2:

Sₙ = n/2 × [2×1 + (n – 1)×2]
Sₙ = n/2 × [2 + 2n – 2]
Sₙ = n/2 × 2n
Sₙ = n²

Answer: Sₙ = n²

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Q10

Show that a1 , a2 , . . ., an , . . . form an AP where an is defined as below :

(i) an = 3 + 4n

Solution:

We are given a general formula for the n-th term of a sequence:

aₙ = 3 + 4n

Step 1: Show that the sequence is an AP

We find the first few terms:

• a₁ = 3 + 4×1 = 7

• a₂ = 3 + 4×2 = 11

• a₃ = 3 + 4×3 = 15

• a₄ = 3 + 4×4 = 19

The terms are: 7, 11, 15, 19, ...

The difference between consecutive terms is:

• a₂ – a₁ = 11 – 7 = 4

• a₃ – a₂ = 15 – 11 = 4

• a₄ – a₃ = 19 – 15 = 4

Since the difference is constant, the sequence is an Arithmetic Progression (AP) with:

• First term a = 7

• Common difference d = 4

Step 2: Find the sum of the first 15 terms

We use the formula:
Sₙ = n/2 × [2a + (n – 1) × d]

Substitute the values:
a = 7, d = 4, n = 15

S₁₅ = 15/2 × [2×7 + (15 – 1)×4]
S₁₅ = 15/2 × [14 + 56]
S₁₅ = 15/2 × 70
S₁₅ = (15 × 70) / 2 = 1050 / 2 = 525

Answer:

Yes, the sequence aₙ = 3 + 4n forms an AP.
The sum of the first 15 terms is S₁₅ = 525.

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Q10

(ii) an = 9 – 5n

Solution:

We are given the general term of a sequence:

aₙ = 9 – 5n

Step 1: Show that it forms an AP

Find the first few terms:

• a₁ = 9 – 5×1 = 4

• a₂ = 9 – 5×2 = –1

• a₃ = 9 – 5×3 = –6

• a₄ = 9 – 5×4 = –11

So the sequence is: 4, –1, –6, –11, ...

Check the common difference:

• a₂ – a₁ = –1 – 4 = –5

• a₃ – a₂ = –6 – (–1) = –5

• a₄ – a₃ = –11 – (–6) = –5

Since the difference is constant, it is an Arithmetic Progression (AP) with:

• First term a = 4

• Common difference d = –5

Step 2: Find the sum of the first 15 terms

Use the sum formula:
Sₙ = n/2 × [2a + (n – 1) × d]

Substitute values:
a = 4, d = –5, n = 15

S₁₅ = 15/2 × [2×4 + (15 – 1)×(–5)]
S₁₅ = 15/2 × [8 + (–70)]
S₁₅ = 15/2 × (–62)
S₁₅ = (15 × –62) / 2 = –930 / 2 = –465

Answer:

The sequence aₙ = 9 – 5n forms an AP.
The sum of the first 15 terms is –465.

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Q11

If the sum of the first n terms of an AP is 4n – n2 , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution:

We are given the sum of the first n terms of an arithmetic progression (AP) as:

Sₙ = 4n – n²

We are to find:

1. First term (a₁ = S₁)

2. Sum of first two terms (S₂)

3. Second term (a₂ = S₂ – S₁)

4. Third term (a₃ = S₃ – S₂)

5. Tenth term (a₁₀ = S₁₀ – S₉)

6. General term (aₙ = Sₙ – Sₙ₋₁)

Step-by-step:

1. First term (S₁):

S₁ = 4(1) – (1)² = 4 – 1 = 3

2. Sum of first two terms (S₂):

S₂ = 4(2) – (2)² = 8 – 4 = 4

3. Second term:

a₂ = S₂ – S₁ = 4 – 3 = 1

4. Third term:

S₃ = 4(3) – (3)² = 12 – 9 = 3
a₃ = S₃ – S₂ = 3 – 4 = –1

5. Tenth term:

S₁₀ = 4(10) – (10)² = 40 – 100 = –60
S₉ = 4(9) – (9)² = 36 – 81 = –45
a₁₀ = S₁₀ – S₉ = –60 – (–45) = –15

6. General n-th term:

aₙ = Sₙ – Sₙ₋₁
= [4n – n²] – [4(n – 1) – (n – 1)²]
= 4n – n² – [4n – 4 – (n² – 2n + 1)]
= 4n – n² – [4n – 4 – n² + 2n – 1]
= 4n – n² – (4n – 4 – n² + 2n – 1)
= 4n – n² – (4n + 2n – n² – 5)
= 4n – n² – (6n – n² – 5)
= 4n – n² – 6n + n² + 5
= –2n + 5

So, aₙ = 5 – 2n

Answers:

• First term (S₁) = 3

• Sum of first two terms (S₂) = 4

• Second term (a₂) = 1

• Third term (a₃) = –1

• Tenth term (a₁₀) = –15

• General term: aₙ = 5 – 2n

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Q12

Find the sum of the first 40 positive integers divisible by 6.

Solution:

To find the sum of the first 40 positive integers divisible by 6, follow these steps:

Step 1: List the sequence

The numbers divisible by 6 form an Arithmetic Progression (AP):
6, 12, 18, 24, ..., 6 × 40 = 240

• First term, a = 6

• Common difference, d = 6

• Number of terms, n = 40

Step 2: Use the sum formula

Sₙ = n/2 × [2a + (n – 1) × d]

Substitute the values:

S₄₀ = 40/2 × [2×6 + (40 – 1)×6]
S₄₀ = 20 × [12 + 39×6]
S₄₀ = 20 × [12 + 234]
S₄₀ = 20 × 246 = 4920

Answer: 4920

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Q13

Find the sum of the first 15 multiples of 8.

Solution:

Step 1: List the AP

The first 15 multiples of 8 form an AP:
8, 16, 24, 32, ..., 8 × 15 = 120

• First term a = 8

• Common difference d = 8

• Number of terms n = 15

Step 2: Use the sum formula

Sₙ = n/2 × [2a + (n – 1) × d]

Substitute the values:

S₁₅ = 15/2 × [2×8 + (15 – 1)×8]
S₁₅ = 15/2 × [16 + 112]
S₁₅ = 15/2 × 128
S₁₅ = (15 × 128) / 2 = 1920 / 2 = 960

Answer: 960

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Q14

Find the sum of the odd numbers between 0 and 50.

Solution:

Step 1: Understand the sequence

The odd numbers between 0 and 50 are:
1, 3, 5, 7, ..., 49

This is an Arithmetic Progression (AP) where:

• First term (a) = 1

• Common difference (d) = 2

• Last term (l) = 49

We need to find how many terms (n) are there, and then find the sum.

Step 2: Find the number of terms (n)

We use the formula for the n-th term of an AP:
aₙ = a + (n – 1) × d

Here,

• aₙ = 49 (last term)

• a = 1

• d = 2

Substitute into the formula:

49 = 1 + (n – 1) × 2

Step-by-step solving:

1. Subtract 1 from both sides:
49 – 1 = (n – 1) × 2
48 = (n – 1) × 2

2. Divide both sides by 2:
48 ÷ 2 = n – 1
24 = n – 1

3. Add 1 to both sides:
24 + 1 = n
⇒ n = 25

So, there are 25 odd numbers between 0 and 50.

Step 3: Use the sum formula

Now, use the formula for the sum of n terms of an AP:

Sₙ = n/2 × (a + l)

Substitute the values:

S₂₅ = 25/2 × (1 + 49)
S₂₅ = 25/2 × 50
S₂₅ = (25 × 50) ÷ 2
S₂₅ = 1250 ÷ 2
S₂₅ = 625

Final Answer: 625

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Q15

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution:

We are given that the penalty increases daily in an arithmetic pattern:

• 1st day penalty = ₹200

• 2nd day penalty = ₹250

• 3rd day penalty = ₹300

• ...

• 30th day penalty = ?

So the penalties form an Arithmetic Progression (AP).

Step 1: Identify values

• First term, a = 200

• Common difference, d = 50

• Number of terms (days delayed), n = 30

We are to find the sum of penalties for 30 days, i.e., S₃₀

Step 2: Use the sum formula

Sₙ = n/2 × [2a + (n – 1) × d]

Substitute values:

S₃₀ = 30/2 × [2×200 + (30 – 1) × 50]
S₃₀ = 15 × [400 + 29 × 50]
S₃₀ = 15 × [400 + 1450]
S₃₀ = 15 × 1850
S₃₀ = ₹27,750

Answer: ₹27,750

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Q16

A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Solution:

We are given:

• Total prize money = ₹700

• Number of prizes = 7

• Each prize is ₹20 less than the previous one

This means the prizes form a decreasing Arithmetic Progression (AP).

Step 1: Let the first prize be ₹x

Then the 7 prizes will be:

x, x – 20, x – 40, x – 60, x – 80, x – 100, x – 120

Step 2: Find the total sum

Total sum of prizes =
x + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = ₹700

Add all terms:

= 7x – (20 + 40 + 60 + 80 + 100 + 120)
= 7x – 420

So,

7x – 420 = 700

Step 3: Solve for x

Add 420 on both sides:

7x = 700 + 420
7x = 1120

Now divide both sides by 7:

x = 1120 ÷ 7
x = 160

Step 4: Find all prize values

Now use x = 160:

• 1st prize: ₹160

• 2nd prize: ₹140

• 3rd prize: ₹120

• 4th prize: ₹100

• 5th prize: ₹80

• 6th prize: ₹60

• 7th prize: ₹40

Answer: Values of the prizes (in ₹) are:
₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40

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Q17

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

We are told:

• There are 12 classes: Class I to Class XII

• Each class number = number of trees each section of that class will plant

• There are 3 sections per class

Step 1: Understand the pattern

For Class I:
Each section plants 1 tree, and there are 3 sections ⇒ 3 × 1 = 3 trees

For Class II:
Each section plants 2 trees, and there are 3 sections ⇒ 3 × 2 = 6 trees

For Class III:
3 × 3 = 9 trees
...
For Class XII:
3 × 12 = 36 trees

So, the total trees planted will be:

3 × (1 + 2 + 3 + ... + 12)

Step 2: Use the formula for sum of first n natural numbers

Sum = n(n + 1)/2

Here, n = 12

Sum = 12 × (12 + 1) ÷ 2 = 12 × 13 ÷ 2 = 78

Now multiply by 3:

Total trees = 3 × 78 = 234

Answer: 234 trees

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Q18

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig 5.4 (refer book). What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take p = 22/7 )

Solution:

We are given a spiral made up of 13 semicircles, with the radii increasing by 0.5 cm for each next semicircle. The centers alternate between A and B, and the sequence of radii is:

r₁ = 0.5 cm, r₂ = 1.0 cm, r₃ = 1.5 cm, ..., r₁₃ = 6.5 cm

Each semicircle has length:

l = (π × r)  [since the length of a semicircle is (½ × 2πr) = πr]

Step 1: List lengths of each semicircle

• l₁ = π × 0.5

• l₂ = π × 1.0

• l₃ = π × 1.5

• l₄ = π × 2.0

• ...

• l₁₃ = π × 6.5

So the total length of the spiral =
l₁ + l₂ + l₃ + ... + l₁₃ = π × (0.5 + 1.0 + 1.5 + ... + 6.5)

Step 2: Use sum of an AP

This is an arithmetic progression where:

• First term a = 0.5

• Last term = 6.5

• Number of terms = 13

Use the formula:
Sum = n/2 × (first term + last term)
= 13/2 × (0.5 + 6.5)
= 13/2 × 7 = 91/2

Now, multiply by π:

Total length = π × 91/2

Take π = 22/7:

= (22/7) × (91/2)
= (22 × 91) ÷ (7 × 2)
= 2002 ÷ 14
= 143 cm

Answer: 143 cm

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Q19

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig. 5.5 (refer book)). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

We are given that 200 logs are stacked with:

• Bottom row = 20 logs

• Each next row has 1 less log

• Total logs = 200

• We need to find the number of rows (n) and the logs in the top row (aₙ)

This is an arithmetic progression (AP) where:

• First term a = 20

• Common difference d = –1

• Sum S = 200

• Formula:
  S = n/2 × [2a + (n – 1)d]

Step 1: Use the sum formula

S = 200, a = 20, d = –1

Plug into the formula:

200 = n/2 × [2×20 + (n – 1)(–1)]
200 = n/2 × [40 – (n – 1)]
200 = n/2 × (41 – n)

Multiply both sides by 2:

400 = n × (41 – n)
400 = 41n – n²
n² – 41n + 400 = 0

Step 2: Solve the quadratic

n² – 41n + 400 = 0
Solve using factorization or quadratic formula:

Factors of 400 that sum to 41: 25 and 16
n² – 25n – 16n + 400 = 0
(n – 25)(n – 16) = 0
n = 25 or n = 16

Check feasibility:

• If n = 25: Top row = aₙ = a + (n – 1)d = 20 + 24(–1) = –4 ❌ Not possible

• If n = 16: a₁₆ = 20 + 15(–1) = 5 Valid

Answer:

• Number of rows = 16

• Logs in the top row = 5

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Q20

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig. 5.6(refer book)). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

[Hint : To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Solution:

We are given:

• A bucket is placed at the starting point.

• The first potato is 5 m away from the bucket.

• The next 9 potatoes are placed 3 m apart in a line.

• So there are 10 potatoes in total.

• The runner starts from the bucket, goes to each potato one by one, brings it back to the bucket, and repeats until all are picked.

Step-by-step calculation:

1. Distances from the bucket to each potato:

• Potato 1: 5 m

• Potato 2: 5 + 3 = 8 m

• Potato 3: 5 + 6 = 11 m

• Potato 4: 5 + 9 = 14 m

• Potato 5: 5 + 12 = 17 m

• Potato 6: 5 + 15 = 20 m

• Potato 7: 5 + 18 = 23 m

• Potato 8: 5 + 21 = 26 m

• Potato 9: 5 + 24 = 29 m

• Potato 10: 5 + 27 = 32 m

So the distances are: 5, 8, 11, 14, 17, 20, 23, 26, 29, 32

This forms an arithmetic progression with:

• First term = 5

• Common difference = 3

• Number of terms = 10

2. Total one-way distance = sum of all distances = 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32

Sum = (10 / 2) × (first term + last term)
Sum = 5 × (5 + 32) = 5 × 37 = 185

Since the runner goes to each potato and comes back, the distance is doubled:

Total distance run = 2 × 185 = 370 metres

Answer: 370 metres

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Exercise 5.4 (optional)

Q1

Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]

Solution:

We are given the AP: 121, 117, 113, ...

Step 1: Identify the first term and common difference

• First term (a) = 121

• Common difference (d) = 117 - 121 = -4

Let the nth term be the first negative term.
So, we want:
a + (n - 1) × d < 0

Step 2: Use the formula and solve

121 + (n - 1) × (-4) < 0
121 - 4(n - 1) < 0
121 - 4n + 4 < 0
125 - 4n < 0
-4n < -125
n > 125 ÷ 4
n > 31.25

So the smallest integer greater than 31.25 is 32

Answer: 32nd term is the first negative term.

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Q2

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Solution:

Given:

• The sum of the 3rd and 7th terms of an AP is 6:
  a₃ + a₇ = 6

• Their product is 8:
  a₃ × a₇ = 8

We are to find the sum of the first 16 terms of the AP, that is S₁₆.

Step 1: Use general formula of the nth term

The nth term of an AP is:
aₙ = a + (n − 1)d

So,

• a₃ = a + 2d

• a₇ = a + 6d

Step 2: Use the two given equations

From the sum:
a₃ + a₇ = (a + 2d) + (a + 6d) = 2a + 8d = 6
→ Divide both sides by 2:
a + 4d = 3 → (Equation 1)

From the product:
a₃ × a₇ = (a + 2d)(a + 6d) = 8 → (Equation 2)

Step 3: Substitute Equation 1 into Equation 2

From Equation 1:
a = 3 − 4d

Now substitute this into Equation 2:

(a + 2d)(a + 6d) = 8
→ (3 − 4d + 2d)(3 − 4d + 6d) = 8
→ (3 − 2d)(3 + 2d) = 8

Now use identity: (x − y)(x + y) = x² − y²

→ (3 − 2d)(3 + 2d) = 9 − 4d² = 8
→ 9 − 4d² = 8
→ 4d² = 1
→ d² = 1/4
→ d = 1/2 or d = −1/2

Step 4: Find values of a

From Equation 1:
a = 3 − 4d

If d = 1/2, then
a = 3 − 4(1/2) = 3 − 2 = 1

If d = −1/2, then
a = 3 − 4(−1/2) = 3 + 2 = 5

So, we have two possible APs:

• Case 1: a = 1, d = 1/2

• Case 2: a = 5, d = −1/2

Step 5: Use the sum formula

Sₙ = n/2 × [2a + (n − 1)d]

We are asked to find S₁₆

Case 1: a = 1, d = 1/2

S₁₆ = 16/2 × [2 × 1 + 15 × 1/2]
= 8 × [2 + 7.5]
= 8 × 9.5
= 76

Case 2: a = 5, d = −1/2

S₁₆ = 16/2 × [2 × 5 + 15 × (−1/2)]
= 8 × [10 − 7.5]
= 8 × 2.5
= 20

Answer: S₁₆ = 20, 76

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Q3

A ladder has rungs 25 cm apart. (see Fig. 5.7 (refer book)). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2(1/2) m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = (250/25) +1]

Solution:

Given:

• Distance between rungs = 25 cm

• Bottom rung length = 45 cm

• Top rung length = 25 cm

• Distance between top and bottom rung = 2½ m = 250 cm

Step 1: Number of rungs
Number of rungs = (250 ÷ 25) + 1 = 10 + 1 = 11

Step 2: The lengths of the rungs form an arithmetic progression (AP)
First term (a) = 45 cm
Last term (l) = 25 cm
Number of terms (n) = 11

Step 3: Use AP sum formula
Sum = n × (a + l) ÷ 2
= 11 × (45 + 25) ÷ 2
= 11 × 70 ÷ 2
= 11 × 35
= 385 cm

Answer: The total length of wood required is 385 cm.

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Q4

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

[Hint : Sx – 1 = S49 – Sx ]

Solution:

We are given:

• House numbers: 1 to 49

• We are to find a house number x such that:
  Sum of house numbers before x = Sum of house numbers after x

Step-by-step solution:

Let total sum of house numbers from 1 to 49 be:

S₄₉ = (49 × 50) / 2 = 1225

Let the sum of house numbers before house number x be:

Sₓ₋₁ = (x – 1) × x / 2

Then the sum of numbers after house number x is:

S₄₉ – Sₓ = 1225 – [x × (x + 1) / 2]

From the question:
Sₓ₋₁ = S₄₉ – Sₓ

So we write:

(x – 1) × x / 2 = 1225 – [x × (x + 1) / 2]

Multiply all terms by 2:

(x – 1) × x = 2450 – x × (x + 1)

Expand both sides:

x² – x = 2450 – (x² + x)
x² – x = 2450 – x² – x

Bring all terms to one side:

x² – x – 2450 + x² + x = 0
2x² – 2450 = 0
2x² = 2450
x² = 1225
x = √1225 = 35

Answer: x = 35

This means the 35th house is the one where the sum of house numbers before and after it is equal.

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Q5

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see Fig. 5.8 (refer book)). Calculate the total volume of concrete required to build the terrace.

[Hint : Volume of concrete required to build the first step = 1/4 ´ 1/2 ´ 50 m3 ]

Solution:

Given:

• Number of steps = 15

• Rise of each step = 1/4 m

• Tread of each step = 1/2 m

• Length of each step = 50 m

Step-by-step solution:

Volume of one step
= rise × tread × length
= 1/4 × 1/2 × 50
= 6.25 m³

Volume of 15 steps
= 15 × 6.25
= 93.75 m³

But this method doesn't give the total correctly, as the steps form a triangular prism. So we calculate the volume of the full prism.

Total height = 15 × 1/4 = 3.75 m

Total base = 15 × 1/2 = 7.5 m

Area of triangle face
= 1/2 × base × height
= 1/2 × 7.5 × 3.75
= 14.0625 m²

Volume of the prism
= Area × length
= 14.0625 × 50
= 703.125 m³

Rounded off:

Answer: 750 m³