Question

If the sum of $8$ terms of an A.P. is $64$ and the sum of $19$ terms is $361$, find the sum of $n$ terms.

Answer 1

Sum of $1^{st}$ $8$ terms of an A.P $=64$

Sum of first $19$ terms of an A.P$=361$

Let a, d are first term and common difference of an A.P, respectively.

$S_8=64;S_{19}=361$

$S_8=(2a+7d)\times 4$ $(\because S_n=\frac{(2a+(n-1\left)d\right)\times n}{2})$

$s_{19}=(2a+18d)\times \frac{19}{2}$

$(2a+7d)\times {\text{/}4}_1={\text{/}64}_{16}$

$2a+7d=16$ …….$\left(1\right)$

$(2a+18d)\times \frac{{\text{/}19}^1}{2}={\text{/}361}^{19}$

$2a+18d=38$ ……$\left(2\right)$

Subtract $\left(1\right)$ from $\left(2\right)$

${\text{/}11}_{1}d={\text{/}22}_2$

$d=2$

Substitute in $\left(1\right)$

$2a+14=16\Rightarrow a=1$

Sum of 'n' terms $=S_n$

$S_n=\left(\frac{\text{/}2+(n-1)\text{/}d}{\text{/}2}\right)n$

$=[1+(n-1\left)\right]=n(\because d=2)$

$=n^2$.