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

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.

Solution:

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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