Question

How many terms have to be considered for getting the sum $5740$ in the A.P.:

$7,14,21,.....$

Answer 1

The given A.P. is $7,14,21,....$
Also, $S_n=5740$
Here, $a=7,d=14-7=7$
Now, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$5740=\frac{n}{2}\left[2\right(7)+(n-1\left)7\right]$
$5740\times 2=n[14+7n-7]$
$\therefore 11480=n(7n+7)$
$\therefore 7n^2+7n-11480=0$
$\therefore 7[n^2+n-1640]=0$
$\therefore n^2+n-1640=0$
$\therefore n^2+41n-40n-1640=0$
$\therefore n(n+41)-40(n+41)=0$
$\therefore (n+41)(n-40)=0$
$\therefore n+41=0$ or $n-40=0$
$\therefore n=-41$ or $n=40$
But, $n$ cannot be negative
$\therefore n\ne -41$
$\therefore n=40$
$\therefore$ The number of terms to be considered are $40$.