Question

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

Answer 1

Given,

Sum of 7 terms of an A.P. is 49

⟹ S₇ = 49

And, sum of 17 terms of an A.P. is 289

⟹ S₁₇ = 289

Let the first term of the A.P be a and common difference as d.

And, we know that the sum of n terms of an A.P is

S_n =\(\frac{ n}{2}\)[2a + (n − 1)d]

So,

S₇ = 49 = \(\frac{7}{2}\)[2a + (7 – 1)d]

= \(\frac{7}{2}\) [2a + 6d]

= 7[a + 3d]

⟹ 7a + 21d = 49

a + 3d = 7 ….. (i)

Similarly,

S₁₇ = \(\frac{17}{2}\)[2a + (17 – 1)d]

= \(\frac{17}{2}\) [2a + 16d]

= 17[a + 8d]

⟹ 17[a + 8d] = 289

a + 8d = 17 ….. (ii)

Now, subtracting (i) from (ii), we have

a + 8d – (a + 3d) = 17 – 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 – 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_n  = \(\frac{n}{2}\)[2(1) + (n − 1)(2)]

= \(\frac{n}{2}\)[2 + 2n – 2]

= \(\frac{n}{2}\)[2n]

= n²

Therefore, the sum of n terms of the A.P is given by n².