Question

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

Answer 1

We have,

Let first term is an A.P.

$a=-14$

Fifth term is $a_5=a+4d=2$

Put $a=-14$ and we get,

$a+4d=2$

$-14+4d=2$

$4d=2+14$

$4d=16$

$d=4$

Find the value of n

Whose sum is $40$.

Then,

We know that,

$S_n=\frac{n}{2}(2a+(n-1)d)$

$\Rightarrow 40=\frac{n}{2}(2\times (-14)+(n-1)\times 4)$

$\Rightarrow 80=n(-28+4n-4)$

$\Rightarrow 80=n(-32+4n)$

$\Rightarrow 4n(n-8)=80$

$\Rightarrow n(n-8)=20$

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

$\Rightarrow n^2-(10-2)n-20=0$

$\Rightarrow n^2-10n+2n-20=0$

$\Rightarrow n(n-10)+2(n-10)=0$

$\Rightarrow (n-10)(n+2)=0$

For

$n+2=0$

$n=-2$

It is not possible (negative)

For,

$n-10=0$

$n=10$

Hence, this is the answer.