Question

In the A.P. 7, 14, 21, ... How many terms are to be considered for getting sum 5740.

• A. 40
• B. 50
• C. 14
• D. 51

Answer 1

The correct option is A 40

Given AP is 7,14,21.....
Since, first term $a=7$ and common difference $d=7$
Suppose no. of terms in given AP= $n$
Sum of $n$ terms $S_n=5740$
Since $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 5740=\frac{n}{2}[2\times 7+(n-1\left)7\right]$
$\Rightarrow 5740=\frac{n}{2}[14+7n-7]$
$\Rightarrow 5740=\frac{n}{2}[7+7n]$
$\Rightarrow 5740=\frac{n}{2}\times 7[1+n]$
$\Rightarrow 5740\times 2=7n[1+n]$
$\Rightarrow 11480=7n[1+n]$
$\Rightarrow \frac{11480}{7}=n[n+1]$
$\Rightarrow 1640=n^2+n$
$\Rightarrow n^2+n-1640=0$
$\Rightarrow (n+41)(n-40)=0$
$\Rightarrow n=-41,n=40$
no. of terms $n$ can't be negative.
$\therefore n=40$