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:

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 nth term of an AP is aₙ = a + (n - 1)d

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

Given,

• Sum of first 7 terms, S₇ = 49
• Sum of first 17 terms, S₁₇ = 289

We know that sum of n terms of AP is Sₙ = n/2 [2a + (n - 1) d]

S₇ = 7/2 [2a + (7 - 1)d]

49 = 7/2 [2a + 6d]

a + 3d = 7 ... (i)

S₁₇ = 17/2 [2a + (17 - 1) d]

289 = 17/2 [2a + 16d]

a + 8d = 17 ... (ii)

Subtracting equation (i) from equation (ii),

a + 8d - (a + 3d) = 17 - 7

5d = 10

d = 2

From equation (i),

7 = a + 3 × 2

7 = a + 6

a = 1

Sₙ = n/2 [2a + (n - 1) d]

= n/2 [2 × 1 + (n - 1) 2]

= n/2 [2 + 2n - 2]

= n/2 × 2n

= n²

☛ Check: NCERT Solutions for Class 10 Maths Chapter 5

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Video Solution:

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

Class 10 Maths NCERT Solutions Chapter 5 Exercise 5.3 Question 9

Summary:

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, the sum of first n terms is equal to n².

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