Question

The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by $10\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, find the number of terms and the series.

Answer 1

Let total number of terms be 2n.
According to question, we have:

$$\begin{gathered} a_1+a_3+...+a_{2n-1}=24...\left(1\right) \\ {a}_2+a_4+...+a_{2n}=30...\left(2\right) \\ \mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}: \\ \left(d+d+...+\mathrm{upto}n\mathrm{terms}\right)=6 \\ \Rightarrow nd=6...\left(3\right) \\ \mathrm{Given}: \\ {a}_{2n}=a_1+\frac{21}{2} \\ \Rightarrow a_{2n}-a_1=\frac{21}{2} \\ \Rightarrow a+(2n-1)d-a=\frac{21}{2}[\because a_{2n}=a+(2n-1)d,a_1=a] \\ \Rightarrow 2nd-d=\frac{21}{2} \\ \Rightarrow 2\times 6-d=\frac{21}{2}\left(\mathrm{From}\left(3\right)\right) \\ \Rightarrow d=\frac{3}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{in}\left(3\right),\mathrm{we}\mathrm{get}: \\ n=4 \\ \Rightarrow 2n=8 \\ \mathrm{Thus},\mathrm{there}\mathrm{are}8\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{progression}. \end{gathered}$$

$$\begin{gathered} \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{first}\mathrm{term}: \\ {a}_2+a_4+...+a_{2n}=30 \\ \Rightarrow (a+d)+(a+3d)+...+[a+(2n-1\left)d\right]=30 \\ \Rightarrow \frac{n}{2}\left[\left(a+d\right)+a+(2n-1)d\right]=30 \\ \mathrm{Putting}n=4\mathrm{and}d=\frac{3}{2},\mathrm{we}\mathrm{get}: \\ a=\frac{3}{2} \\ \mathrm{So},\mathrm{the}\mathrm{series}\mathrm{will}\mathrm{be}1\frac{1}{2},3,4\frac{1}{2}... \end{gathered}$$