Question

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer 1

Given:

$$\begin{gathered} a_6=19 \\ \Rightarrow a+\left(6-1\right)d=19\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+5d=19..\left(1\right) \\ \mathrm{And},a_{17}=41 \\ \Rightarrow a+\left(17-1\right)d=41\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+16d=41..\left(2\right) \\ \mathrm{Solving}\mathrm{the}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}, \\ 16d-5d=41-19 \\ \Rightarrow 11d=22 \\ \Rightarrow d=2 \\ \mathrm{Putting}d=2\mathrm{in}\mathrm{the}\mathrm{eqn}\left(1\right),\mathrm{we}\mathrm{get}: \\ a+5\times 2=19 \\ \Rightarrow a=19-10 \\ \Rightarrow a=9 \end{gathered}$$


We know:
$$\begin{gathered} a_{40}=a+\left(40-1\right)d\left[a_n=a+\left(n-1\right)d\right] \\ =a+39d \\ =9+39\times 2 \\ =9+78=87 \end{gathered}$$