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 101/2, find the number of terms and the series.

Answer 1

Let total no. of terms be 2n
A.T.Q
$\Rightarrow T_1+T_3+T_5+\cdots T_{2n-1}=24\cdots \left(1\right)$
$\Rightarrow T_2+T_4+T_6+\cdots T_{2n=30}\cdots \left(2\right)$
Subtract 1 from 2
$(d+d+d+d+d\cdots )=6$
$nd=6$
$T_{2n}=T_1+\frac{21}{2}$
$\Rightarrow T_{2n}-a=\frac{21}{2}$
$\Rightarrow a+(2n-1)d-a=\frac{21}{2}$
$\Rightarrow 2nd-d=\frac{21}{2}$
$\Rightarrow 2\left(6\right)-d=\frac{21}{2}$
$\Rightarrow 12-\frac{21}{2}=d\Rightarrow d=\frac{3}{2}$
$andnd=6$
$\Rightarrow n\left(\frac{3}{2}\right)=6$
$\Rightarrow n=4$
$\Rightarrow 2n=8$
To find the value of first term, $T_2+T_4+T_6+\cdots \cdots T_{2n}=30$
$(a+d)+(a+3d)+(a+5d)\cdots \cdots [a+(2n-1\left)d\right]=30$
$\Rightarrow \frac{n}{2}\left[\right(a+d)+\{a+(2n-1)\left\}\right]=30$
By putting n = 4 and $d=\frac{3}{2},wegeta=\frac{3}{2}$
So, the series will be $1\frac{1}{2},3,4\frac{1}{2}\cdots$