Question

The number of terms of an A.P. is even; the sum of the odd terms is 24, and of the even terms is 30, and the last term exceeds the first by $\frac{21}{2}$, then the number of terms in the series is

• A. 3
• B. 4
• C. 6
• D. 5

Answer 1

The correct option is

C

4

Let the series have 2n terms and the series is a, $a+d,a+2d,.....a+(2n-1)$d.

According to the given conditions, we have

$[a+(a+2d)+(a+4d)+.....+(a+(2n-2)d\left)\right]=24$

$\Rightarrow \frac{n}{2}\left[2a\right(n-1\left)2d\right]=24$

$\Rightarrow n[a+(n-1\left)d\right]=24$.....(1)

Also,

$\left[\right(a+d)+(a+3d)+....+(a+(2n-1)d\left)\right]=30$

$\Rightarrow \frac{n}{2}\left[2a\right(a+d)+(n-1\left)2d\right]=30$

$\Rightarrow n\left[\right(a+d)+(n-1\left)d\right]=30$......(2)

Also, the last term exceeds the first by $\frac{21}{2}.$ Therefore,

$a+(2n-1)d-a=\frac{21}{2}$

$\Rightarrow (2n-1)d=\frac{21}{2}$......(3)

Now, subtracting $\left(1\right)$ from $\left(2\right)$,

$nd=6$...(4)

Dividing $\left(3\right)$by$\left(4\right)$, we get

$\frac{2n-1}{n}=\frac{21}{12}$

$\Rightarrow n=4$