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.
Let total number of terms be 2n.
According to question, we have:
$a_{1}+a_{3}+\ldots+a_{2 n-1}=24 \ldots(1)$
$a_{2}+a_{4}+\ldots+a_{2 n}=30 \quad \ldots(2)$
Subtracting (1) from (2), we get:
$\Rightarrow n d=6$ ....(3)
Given:
$a_{2 n}=a_{1}+\frac{21}{2}$
$\Rightarrow a_{2 n}-a_{1}=\frac{21}{2}$
$\Rightarrow a+(2 n-1) d-a=\frac{21}{2} \quad\left[\because a_{2 n}=a+(2 n-1) d, a_{1}=a\right]$
$\Rightarrow 2 n d-d=\frac{21}{2}$
$\Rightarrow 2 \times 6-d=\frac{21}{2} \quad(\operatorname{From}(3))$
$\Rightarrow d=\frac{3}{2}$
Putting the value in $(3)$, we get:
$n=4$
$\Rightarrow 2 n=8$
Thus, there are 8 terms in the progression.
To find the value of the first term:
$a_{2}+a_{4}+\ldots+a_{2 n}=30$
$\Rightarrow(a+d)+(a+3 d)+\ldots+[a+(2 n-1) d]=30$
$\Rightarrow \frac{n}{2}[(a+d)+a+(2 n-1) d]=30$
Putting $n=4$ and $d=\frac{3}{2}$, we get:
$a=\frac{3}{2}$
So, the series will be $1 \frac{1}{2}, 3,4 \frac{1}{2} \ldots$
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