How many terms are there in the A.P. whose first and fifth terms are $-14$ and $2$ respectively and the sum of the terms is $40$?

Given:

The first and fifth terms of an A.P. are $-14$ and $2$ respectively and the sum of the terms is $40$.
To do:

We have to find the number of terms in the A.P.
Solution:

First term of the A.P. \( =-14 \) and fifth term \( =2 \)

Sum of terms \( =40 \)
Let \( n \) be the number of terms.

\( a_{5}=2 \)
\( \Rightarrow a_{5}=a+(5-1) d \)
\( \Rightarrow 2=-14+4 d \)

\( \Rightarrow 4 d=14+2=16 \)
\( \Rightarrow d=\frac{16}{4}=4 \)
We know that,
\( \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d] \)

$40=\frac{n}{2}[2\times(-14)+(n-1)\times4$

$40=\frac{2n}{2}[-14+2n-2]$

$40=n(2n-16)$

$40=2n(n-8)$

$20=n^2-8n$

$n^2-8n-20=0$

\( n(n-10)+2(n-10)=0 \)
\( (n-10)(n+2)=0 \)
\( n-10=0 \) or  \( n+2=0 \)

\( n=10\) or \( n=-2 \) which is not possible
\( \therefore \) \( n=10 \)

The number of terms in the A.P. is 10.