Question

The sum of first $2n$ terms of an AP is $\alpha$ and the sum of next $n$ terms is $\beta$, its common difference is

• A. $\frac{\alpha -2\beta }{3n^2}$
• B. $\frac{2\beta -\alpha }{3n^2}$
• C. $\frac{\alpha -2\beta }{3n}$
• D. $\frac{2\beta -\alpha }{3n}$

Answer 1

Answer: (B)

$\frac{2\beta -\alpha }{3n^2}$

Solution :-

Given, sum of first 2n terms is $\alpha$

Let first term be a

$\therefore S_1=\frac{2n}{2}\{2a+(2n-1\left)d\right\}=\alpha$

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

Sum of first $3_n$ terms of an A.P

$S_2=\frac{3n}{2}\{2a+(3n-1\left)d\right\}=\beta +\alpha$

$\Rightarrow \alpha +\beta =\frac{3n}{2}\{\frac{\alpha }{2}-(2n-1)d+(3n-1\left)d\right\}$

$\Rightarrow \frac{\alpha }{n}+nd=\frac{2(\alpha +\beta )}{3n}$

$\Rightarrow \frac{2(\alpha +\beta )}{3n}=\frac{\alpha }{n}=nd$

$\Rightarrow \frac{2\alpha n+2\beta n-3\alpha n}{3n^2}=nd$

$\Rightarrow \frac{2\beta n-\alpha n}{3n}=nd$

$\frac{2\beta -\alpha }{3n^2}=d$ [common Difference]