Question

16 A.Ms are inserted between 5 and 50. Find the sum of all the A.Ms.

• A. 880
• B. 820
• C. 410
• D. 440

Answer 1

The correct option is D

440

Let the common difference of the resultant AP be d.
$5,5+d,\cdots 50-d,50$
Now, sum of $n$ A.M.'s is given by , $S=\frac{n}{2}[a+l]$
$a=5+d,l=50-d,n=16$
$\Rightarrow S=\frac{16}{2}\left[\right(5+d)+(50-d\left)\right]$
$=8\times 55$
$=440$