Question

If the first, second and last terms of an A.P. are a, b and 2a respectively, then the sum of its terms is ___________.

Answer 1

Let first term of A.P be a
Second term of A.P be b and last term of A.P be given by 2a
Since d = a₂ – a₁ = b – a

$\begin{array}{rcl}{n}^{\mathrm{th}}\mathrm{term}\mathrm{i}.\mathrm{e}{a}_n& =& a+\left(n-1\right)d\\ & =& a+\left(n-1\right)\left(b-a\right)=2a\left(\mathrm{given}\right)\end{array}$
$$\begin{gathered} \Rightarrow a+\left(n-1\right)\left(b-a\right)=2a \\ \Rightarrow \left(n-1\right)\left(b-a\right)=a \\ \mathrm{i}.\mathrm{e}n-1=\frac{a}{b-a}\mathrm{i}.\mathrm{e}n=\frac{a}{b-a}+1=\frac{b}{b-a} \end{gathered}$$
∴ S_n i.e sum of n terms of A.P is
$$\begin{gathered} \frac{n}{2}\left(a_1+a_n\right) \\ =\frac{n}{2}\left(a+2a\right) \\ =\frac{b}{2\left(b-a\right)}\left(3a\right) \end{gathered}$$
∴ Sum of its term = $\frac{3ab}{2\left(b-a\right)}$