Question

The $4^{th}$ term of A.P. is $22$ and ${15}^{th}$ term is $66$. Find the first term and the common difference. Hence find the sum of the series to $8$ terms.

Answer 1

Let the first term be $a$ and the common difference be $d$, then

the fourth term is $a_4=a+3d$ and the fifteenth term is $a_{15}=a+14d$.

By the given conditions, we have:

$a+3d=22,a+14d=66$

Solving the above we get

$\Rightarrow d=4,a=10$

The first term is $10$ and the common difference is $4$.

Thus, the eighth term in the sequence is

$a_8=a+7d=10+7\times 4=38$

The sum of the first $n$ terms in an A.P. is given by the formula

$S=\frac{n}{2}\times (a_1+a_n)$

Thus the sum of the first $8$ terms of the series is:

$\frac{8}{2}(a_1+a_8)=4(10+38)=192$

So, the sum of the first $8$ terms is $192$.