Question

The sums of n$,2n,3n$ terms of an A.P. are $S_1,S_2,S_3$ respectively. Then $S_3=3(S_2-S_1)$.

• A. True
• B. False

Answer 1

The correct option is A True

Here a and d are same as there is same A.P. and N is equal to n, $2n,3n$ for $S_1,S_2$ and $S_3$ respectively:

$3(S_2-S_1)=3\left[\right(2n/2\left)\right\{2a+(2n-1)d\}-(n/2\left)\right\{2a+(n-1)d\left\}\right]$

$=\left(3n/2\right)\left[\right(4a-2a)+(4n-2-n+1\left)d\right]$

$=\left(3n/2\right)[2a+(3n-1\left)d\right]=S_3$

i.e., sum of $3n$ terms.