Question

If the first, second and last terms of an A.P be $a,b,2a$ respectively, then its sum will be

• A. $\frac{ab}{-a+b}$
• B. $\frac{ab}{2(b-a)}$
• C. $\frac{3ab}{2(b-a)}$
• D. $\frac{3ab}{4(b-a)}$

Answer 1

The correct option is C $\frac{3ab}{2(b-a)}$

Given $a_1=a,a_2=b$ $\therefore d=b-a,l=2a$

We use these formulas

nth term $t_n=a+(n-1)d$ and $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Let the number of terms be $n$

$a_n=a_1+(n-1)d$

$\Rightarrow 2a=a+(n-1)(b-a)$

$\Rightarrow \frac{a}{b-a}=n-1$

$\Rightarrow n=\frac{a}{b-a}+1$

$\Rightarrow n=\frac{a+b-a}{b-a}$

$\Rightarrow n=\frac{b}{b-a}$

$\Rightarrow S_n=\frac{b}{2(b-a)}[2a+(\frac{b}{b-a}-1)(b-a\left)\right]$

$\Rightarrow S_n=\frac{b}{2(b-a)}[2a+\left(\frac{b-b+a}{b-a}\right)(b-a\left)\right]$

$\Rightarrow S_n=\frac{b}{2(b-a)}\left(3a\right)$

$\Rightarrow S_n=\frac{3ab}{2(b-a)}$