Question

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

Answer 1

Given,
$a_{2n}=a_1=\frac{21}{2}\Rightarrow a_{2n}-a_1=\frac{21}{2}\Rightarrow a_{2n}-a_1=\frac{21}{2}\Rightarrow a+(2n-1)d-a=\frac{21}{2}[\because a_{2n}=a+(2n-1)d,a_1=a]\Rightarrow 2nd-d=\frac{21}{2}\Rightarrow 2\times 6-d=\frac{21}{2}\text{[From (3)]}\Rightarrow d=\frac{3}{2}$
Putting the value in (3), we get:
n = 4
$\Rightarrow 2n=8$
Thus, there are 8 term in the progression
To find the value of the first term:
$a_2+a_4+\dots \dots +a_{2n}=30\Rightarrow (a+d)+(a+3d)+\dots \dots +[a+(2n-1\left)d\right]=30\Rightarrow \frac{n}{2}\left[\right(a+d)+a+(2n-1\left)d\right]=30$
Putting n = 4 and $d=\frac{3}{2}$, we get:
$a=\frac{3}{2}$
So, the series will be $1\frac{1}{2},3,4\frac{1}{2},\dots \dots$