Question

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Answer 1

Given:

$$\begin{gathered} a_{10=}41 \\ \Rightarrow a+\left(10-1\right)d=41\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+9d=41 \\ \mathrm{And},a_{18}=73 \\ \Rightarrow a+\left(18-1\right)d=73\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a+17d=73 \\ \mathrm{Solving}\mathrm{the}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get}: \\ \Rightarrow 17d-9d=73-41 \\ \Rightarrow 8d=32 \\ \Rightarrow d=4...\left(i\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{in}\mathrm{first}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ a+9\times 4=41 \\ \Rightarrow a+36=41 \\ \Rightarrow a=5...\left(ii\right) \end{gathered}$$

$$\begin{gathered} a_{26}=a+\left(26-1\right)d\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow a_{26}=a+25d \\ \Rightarrow a_{26}=5+25\times 4\left(\mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right) \\ \Rightarrow a_{26}=5+100=105 \end{gathered}$$