Question

If the first, second and last term of an A.P are a, b and 2a respectively, then its sum is
(a) $\frac{ab}{2(b-a)}$

(b) $\frac{ab}{b-a}$

(c) $\frac{3ab}{2(b-a)}$

(d) none of these

Answer 1

(c) $\frac{3ab}{2(b-a)}$

Let the A.P. be a, a+d, a+2d........a+nd.
Here, let d be the common difference and n be the total number of terms.

Given:
$$\begin{gathered} a_1=a, \\ {a}_2=b \\ \Rightarrow a+d=b \\ \Rightarrow d=b-a.....\left(1\right) \\ {a}_n=2a \\ \Rightarrow a+\left(n-1\right)d=2a \\ \Rightarrow \left(n-1\right)d=a \\ \Rightarrow d=\frac{a}{n-1}.....\left(2\right) \end{gathered}$$

From equations $\left(1\right)\mathrm{and}\left(2\right),$ we have:
$$\begin{gathered} \Rightarrow \frac{a}{n-1}=b-a \\ \Rightarrow \frac{a}{b-a}+1=n \\ \Rightarrow \frac{a+b-a}{b-a}=n \\ \Rightarrow \frac{b}{b-a}=n \end{gathered}$$

Now, sum of n terms of an A.P.:
$$\begin{gathered} S=\frac{n}{2}\left\{a+a_n\right\} \\ =\frac{n}{2}\left(3a\right) \\ =\frac{3ab}{2\left(b-a\right)} \end{gathered}$$