Question

The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.

Answer 1

Given:
Let the first term of the A.P. be a and the common difference be d.
$$\begin{gathered} a_4=3a \\ \Rightarrow a+\left(4-1\right)d=3a \\ \Rightarrow a+3d=3a \\ \Rightarrow 3d=2a \\ \Rightarrow a=\frac{3d}{2}...\left(\mathrm{i}\right) \\ \mathrm{And},a_7-2a_3=1 \\ \Rightarrow a+\left(7-1\right)d-2\left[a+\left(3-1\right)d\right]=1 \\ \Rightarrow \mathrm{a}+6d-2(a+2d)=1 \\ \Rightarrow a+6d-2a-4d=1 \\ \Rightarrow -a+2d=1 \\ \Rightarrow -\frac{3d}{2}+2d=1\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{-3d+4d}{2}=1 \\ \Rightarrow \frac{d}{2}=1 \\ \Rightarrow d=2 \end{gathered}$$
$$\begin{gathered} \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ a=\frac{3\times 2}{2} \\ \Rightarrow a=3 \end{gathered}$$