Question

If the sum of first $7$ terms of an Arithmatic Progression is $49$ and that of first $17$ terms is $289$, then find the sum of the first '$n$' terms.

Answer 1

Let first term of AP = a
common difference = d
number of terms = n
Given that sum of first 7 term of AP = $49$
$\frac{n}{2}\left[2a+(n-1)\ast d\right]=49$
$\frac{7}{2}\ast \left[2a+(7-1)\ast d\right]=49$
$\frac{7}{2}\ast \left[2a+6d\right]=49$
$\frac{1}{2}\ast (2a+6d)=7$
$\frac{2}{2}\ast a+3d=7$
$a+3d=\mathrm{7...............1}$
Again given sum of first 17 term of AP = $289$
$\frac{17}{2}\ast 2a+(17-1)\ast d=289$
$\frac{1}{2}\ast 2a+16\ast d=17$ (when 17 and 289 is divided by 17)
$\frac{2}{2}\ast a+8d=17$
$a+8d=17...............2$
After solving equation 1 and 2, we get
$a=1$ and $d=2$
Now sum of n terms of AP = $\frac{n}{2}\ast 2a+(n-1)\ast d$
$\frac{n}{2}\ast 2\ast 1+(n-1)\ast 2$
$\frac{n}{2}\ast (2+2n-2)$
$\frac{n}{2}\ast 2n$
$n^2$
So sum of first n terms of AP = $n^2$