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

Solution:

In the given problem, we need to find the sum of n terms of an A.P. Let us take the first term as a and the common difference as d.

Here, we are given that,

$S_{7}=49$…………..(1)

$S_{17}=289$………….(2)

So, as we know the formula for the sum of n terms of an A.P. is given by,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

So, using the formula for n = 7, we get,

$S_{7}=\frac{7}{2}[2(a)+(7-1)(d)]$

$49=\left(\frac{7}{2}\right)[2 a+(6)(d)]$ (Using 1)

$49=\frac{14 a+42 d}{2}$

$49=7 a+21 d$

Further simplifying for a, we get,

$a=\frac{49-21 d}{7}$

$a=7-3 d$ ……………(3)

Also using the formula for $n=17$ we get

$S_{17}=\frac{17}{2}[2(a)+(17-1)(d)]$

$289=\left(\frac{17}{2}\right)[2 a+(16)(d)]$ (Using 2)

$289=\frac{(17)(2) a+(17)(16) d}{2}$

$289=17 a+136 d$

Further simplifying for a, we get,

$a=\frac{289-136 d}{17}$

$a=17-8 d$ ………..(4)

Subtracting (3) from (4), we get,

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

$0=17-8 d-7+3 d$

$0=10-5 d$

$5 d=10$

$d=2$

Now, to find a, we substitute the value of d in (3),

$a=7-3(2)$

$a=7-6$

$a=1$

Now, using the formula for the sum of n terms of an A.P., we get,

$S_{n}=\frac{n}{2}[2(1)+(n-1)(2)]$

$=\frac{n}{2}[2+2 n-2]$

$=\left(\frac{n}{2}\right)(2 n)$

$=n^{2}$

Therefore, the sum of first $n$ terms for the given A.P. is $S_{n}=n^{2}$.