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# In an AP:

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

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

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀

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

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

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

(viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a

(ix) Given a = 3, n = 8, S = 192 , find d

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

**Solution:**

The sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l], and the n^{th} term of an AP is aₙ = a + (n - 1)d

Here, a is the first term, d is a common difference and n is the number of terms and l is the last term.

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

Given,

- First term, a = 5
- Common difference, d = 3
- n
^{th}term, l = aₙ = 50

aₙ = a + (n - 1)d

50 = 5 + (n - 1)3

45 = (n - 1)3

15 = n - 1

n = 16

Sustitute the value of n, to find sum of the n terms.

Sₙ = n/2 [a + l]

S₁₆_{ }= 16/2 [ 5 + 50]

= 8 × 55

= 440

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

Given,

- First term, a = 7
- 13th term, l = a₁₃ = 35

aₙ = a + (n - 1)d

a₁₃ = a + (13 - 1) d

35 = 7 + 12d

35 - 7 = 12d

d = 28/12

d = 7/3

Sₙ = n/2 [a + l]

S₁₃_{ }= 13/2 [7 + 35]

= 13/2 × 42

= 13 × 21

= 273

(iii) Given a₁₂ = 37, d = 3, find a and S₁₂.

Given,

- 12th term, a₁₂ = 37
- Common Difference, d = 3

aₙ = a + (n - 1)d

a₁₂ = a + (12 - 1) 3

37 = a + 33

a = 4

Substitute the value of a to find the sum of n terms.

Sₙ = n / 2 [a + l]

S₁₂_{ }= 12 / 2 [4 + 37]

= 6 × 41

S₁₂_{ }= 246

(iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Given,

- 3rd term, a₃ = 15
- Sum up to ten terms, S₁₀ = 125.

aₙ = a + (n - 1)d

a₃ = a + (3 - 1) d

15 = a + 2d - - - - - Equation (i)

Sₙ = n/2 [2a + (n - 1) d]

S₁₀ = 10/2 [2a + (10 - 1) d]

125 = 5 [2a + 9d]

25 = 2a + 9d - - - - - Equation (ii)

On multiplying equation (i) by 2, we obtain

30 = 2a + 4d - - - - - Equation (iii)

On subtracting equation (iii) from Equation (ii), we obtain

- 5 = 5d

d = - 1

From equation (i),

15 = a + 2 (- 1)

15 = a - 2

a = 17

a₁₀ = a + (10 - 1) d

a₁₀ = 17 + 9 (- 1)

a₁₀ = 17 - 9

a₁₀ = 8

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

Given,

- Common difference, d = 5
- Sum up to nine terms, S₉ = 75

Sₙ = n/2 [2a + (n - 1) d]

75 = 9/2 [2a + (9 - 1) 5]

75 = 9/2 (2a + 40)

25 = 3(a + 20)

25 = 3a + 60

a = - 35 / 3

We know that nth term of the AP is given by formula aₙ = a + (n - 1)d

a₉ = a + (9 - 1) × 5

= - 35 / 3 + 8 × 5

= - 35 / 3 + 40

= (- 35 + 120) / 3

= 85 / 3

(vi) Given a = 2, d = 8, Sₙ = 90 , find n and a_{n}.

Given,

- First term, a = 2
- Common difference, d = 8
- Sum up to nth terms, Sₙ = 90

Sₙ = n/2 [2a + (n - 1) d]

90 = n/2 [4 + (n - 1) 8]

90 = n [2 + (n - 1)4]

90 = n [2 + 4n - 4]

90 = n [4n - 2]

90 = 4n² - 2n

4n² - 2n - 90 = 0

4n² - 20n + 18n - 90 = 0

4n (n - 5) + 18(n - 5) = 0

(n - 5)(4n + 18) = 0

Either (n - 5) = 0 or (4n + 18) = 0

n = 5 or n = - 9/2

However, n can neither be negative nor fractional.

Therefore, n = 5

aₙ = a + (n - 1) d

a₅ = 2 + (5 - 1)8

a₅ = 2 + 4 × 8

a₅ = 2 + 32

a₅ = 34

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

Given,

- First term, a = 8
- nth term, l = aₙ = 62
- Sum up to nth terms, Sₙ = 210

Sₙ = n / 2 [a + l]

210 = n / 2 [8 + 62]

210 = n × 35

n = 6

We know that n^{th} term of the AP is given by aₙ = a + (n - 1)d

62 = 8 + (6 - 1) d

62 - 8 = 5d

54 = 5d

d = 54/5

(viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a

Given,

- Common difference d = 2
- n
^{th}term, l = aₙ = 4 - Sum up to nth terms, Sₙ = -14

We know that n^{th} term of AP is aₙ = a + (n - 1)d

4 = a + (n - 1)2

4 = a + 2n - 2

a = 6 - 2n .... (1)

Sₙ = n/2 [a + l]

- 14 = n/2 [6 - 2n + 4] [from(1)]

- 14 = n (5 - n)

- 14 = 5n - n²

n² - 5n - 14 = 0

n² - 7n + 2n -14 = 0

n (n - 7) + 2 (n - 7) = 0

(n - 7)(n + 2) = 0

Either n - 7 = 0 or n + 2 = 0

n = 7 or n = - 2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (1), we obtain

a = 6 - 2n

a = 6 - 2 × 7

a = 6 - 14

a = - 8

(ix) Given a = 3, n = 8, S = 192 , find d.

Given,

- First term, a = 3
- Number of terms, n = 8
- Sum up to nth terms, Sₙ = 192

Sₙ = n/2 [2a + (n - 1) d]

192 = 8/2 [2 × 3 + (8 - 1) d]

192 = 4[6 + 7d]

48 = 6 + 7d

42 = 7d

d = 6

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

Given,

- Last term, l = aₙ = 28
- Number of terms, n = 9
- Sum up to nth terms, Sₙ = 144

Sₙ = n/2 [a + l]

144 = 9/2 (a + 28)

32 = a + 28

a = 4

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## In an AP (i) Given a = 5, d = 3, aₙ = 50 , find n and Sₙ. (ii) Given a = 7, a₁₃ = 35 , find d and S₁₃. (iii) Given a₁₂ = 37, d = 3, find a and S₁₂. (iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀. (v) Given d = 5, S₉ = 75 , find a and a₉. (vi) Given a = 2, d = 8, Sₙ = 90 , find n and aₙ. (vii) Given a = 8, aₙ = 62, Sₙ = 210 , find n and d. (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , find n and a. (ix) Given a = 3, n = 8, S = 192 , find d. (x) Given l = 28, S = 144 and there are total 9 terms. Find a

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 3 :

**Summary:**

In an AP (i) Given a = 5, d = 3, aₙ = 50, n = 16, S₁₆_{ }= 400 (ii) Given a = 7, a₁₃ = 35 , d = 7/3, S₁₃_{ }= 273. (iii) Given a₁₂ = 37, d = 3, a = 4, S₁₂_{ }= 246 (iv) Given a₃ = 15, S₁₀ = 125, d = - 1, a₁₀ = 8, (v) Given d = 5, S₉ = 75 , a = - 35 / 3, a₉ = 85/3, (vi) Given a = 2, d = 8, Sₙ = 90 , n = 5, a₅ = 34, (vii) Given a = 8, aₙ = 62, Sₙ = 210 , n = 6, d = 54/5, (viii) Given aₙ = 4, d = 2, Sₙ = - 14 , n = 7, a = - 8, (ix) Given a = 3, n = 8, S = 192 , d = 6, d = 6, (x) Given l = 28, S = 144 and there are total 9 terms, a = 4

**☛ Related Questions:**

- The first and the last term of an AP are 17 and 350 respectively.If the common difference is 9,how many terms are there and what is their sum?
- Find the sum of the first 22 terms of an AP in which d = 7 and 22nd term is 149.
- Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
- 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.

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