The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the A.P. is :

Question

Aicharya · Accepted Answer

$\mathrm{a}_2+\mathrm{a}_4+\ldots+\mathrm{a}_{\mathrm{n}}=30$ ...(1)
$\mathrm{a}_1+\mathrm{a}_3+\ldots+\mathrm{a}_{\mathrm{n}-1}=24$ ...(2)
$(1)-(2)$
$\left(a_2-a_1\right)+\left(a_4-a_3\right) \ldots\left(a_n-a_{n-1}\right)=6$
$\begin{aligned} & \Rightarrow \frac{\mathrm{n}}{2} \mathrm{~d}=6 \Rightarrow \mathrm{nd}=12 \\ & \mathrm{a}_{\mathrm{n}}-\mathrm{a}_1=(\mathrm{n}-1) \mathrm{d}=\frac{21}{2} \\ & \Rightarrow \mathrm{nd}-\mathrm{d}=\frac{21}{2} \Rightarrow 12-\frac{21}{2}=\mathrm{d} \\ & \Rightarrow \mathrm{d}=\frac{3}{2}, \mathrm{n}=8\end{aligned}$
$\begin{aligned}
& \text { Sum of odd terms }=\frac{4}{2}[2 \mathrm{a}+(4-1) 3]=24 \\ & \Rightarrow \mathrm{a}=\frac{3}{2} \\ & \text { A.P. } \Rightarrow \frac{3}{2}, 3, \frac{9}{2}, 6, \frac{15}{2}, 9, \frac{21}{2}, 12
\end{aligned}$
no. of integer terms $=4$ /

The number of terms of an A.P. is even; the sum of all the odd terms is 24 , the sum of all the even terms…

Solution