Question

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

• A. 3
• B. 4
• C. 6
• D. 5

Answer 1

Answer: (B)

Let the series has $2 n$ terms and the series is
$a, a+d, a+2 d, \ldots, a+(2 n-1) d$
According to the given conditions
We have
$[a+(a+2 d)+(a+4 d)+\ldots(a+(2 n-2) d)]=24$
$\Rightarrow \frac{n}{2}[2 a+(n-1) 2 d]=24$
$\Rightarrow n[a+(n-1) d]=24$ ...(1)
Also $[(a+d)+(a+3 d)+\ldots(a+(2 n-1) d)]=30$
$\Rightarrow \frac{n}{2}[2(a+d)+(n-1) 2 d]=30$
$\Rightarrow n[(a+d)+(n-1) d]=30$ ...(2)
Also last term exceeds the first by $21 / 2$
$\Rightarrow a+(2 n-1) d-a=21 / 2$
$\Rightarrow (2 n-1) d=21 / 2$ ...(3)
Now subtracting (1) from (2)
$n d=6$ ...(4)
Dividing $(3)$ by (4)
$\Rightarrow \frac{2 n-1}{n}=\frac{21}{12}$
$\Rightarrow n=4$