Question

The 2nd term of an ap is 38 and 6th term is -22. find the first term , common difference and 10th term of the ap

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{abe}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{and}\mathrm{dbe}\mathrm{the}\mathrm{common}\mathrm{difference}. \\ \mathrm{Now},{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d} \\ \mathrm{Now},{\mathrm{a}}_2=\mathrm{a}+\mathrm{d} \\ \Rightarrow \mathrm{a}+\left(2-1\right)\mathrm{d}=38 \\ \mathrm{Now},\mathrm{a}+\mathrm{d}=38....\left(1\right) \\ \mathrm{Now},{\mathrm{a}}_6=-22 \\ \Rightarrow \mathrm{a}+5\mathrm{d}=-22.....\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 4\mathrm{d}=-60 \\ \Rightarrow \mathrm{d}=-\frac{60}{4}=-15 \\ \mathrm{Now},\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{d}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{a}-15=38 \\ \Rightarrow \mathrm{a}=38+15=53 \\ \mathrm{Now},{\mathrm{a}}_{10}=\mathrm{a}+9\mathrm{d}=53+9\times \left(-15\right)=53-135=-82 \end{gathered}$$