Question

if in an A.P T4+T8= 24 a T6 + T10 =34 then find the four terms of A.p

Answer 1

Let first term of A.P = a
and common difference = d
then n^th term is written as t_n^​ = a + (n - 1) d
It is given that t₄ + t₈ = 24
$$\begin{gathered} \Rightarrow a+3d+a+7d=24 \\ \Rightarrow 2a+10d=24.........\left(1\right) \\ \mathrm{It}\mathrm{is}\mathrm{also}\mathrm{given}\mathrm{that}{t}_6+t_{10}=34 \\ \Rightarrow a+5d+a+9d=34 \\ \Rightarrow 2a+14d=34..........\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right) \\ 4d=10 \\ \Rightarrow d\mathit{}=\frac{10}{4}=\frac{5}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{in}\left(1\right).. \\ 2a+10\times \frac{5}{2}=24 \\ \Rightarrow 2a\mathit{}+25=24 \\ \Rightarrow 2a\mathit{}=\mathit{}24-25 \\ \Rightarrow 2a=-1 \\ \Rightarrow a=\frac{-1}{2} \\ \mathrm{The}\mathrm{required}\mathrm{A}.\mathrm{P}\mathrm{is}\mathrm{written}\mathrm{as} \\ a\mathit{,}\mathit{}a\mathit{+}d\mathit{,}\mathit{}\mathit{}a\mathit{+}\mathit{2}d\mathit{,}\mathit{}\mathit{}a\mathit{+}\mathit{3}d\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.} \\ \Rightarrow \frac{-1}{2},\frac{-1}{2}+\frac{5}{2},\frac{-1}{2}+\frac{10}{2},\frac{-1}{2}+\frac{15}{2}............. \\ \Rightarrow \frac{-1}{2},2,\frac{9}{2},7........... \\ \mathrm{Hence}\mathrm{the}\mathrm{four}\mathrm{terms}\mathrm{of}\mathrm{given}\mathrm{A}.\mathrm{P}\mathrm{are}\frac{-1}{2},2,\frac{9}{2},7 \end{gathered}$$