Question

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

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is B $8$

Let ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{d}^{\prime }$ be the first term and common difference of the series respectively.
Let ${}^{\prime }2n^{\prime }$ be the total terms present in the series.
Now, from the given condition we get
$\frac{n}{2}[2a+(n-1\left)2d\right]=24$ - (for odd terms).............(1)
$\frac{n}{2}[2a+2d+(n-1\left)2d\right]=30$ - (for even terms)...........(2)
Subtract (1) from (2),
$nd=6$.......................(3)
$a+(2n-1)d-a=10.5$.............(4)
$2nd-d=10.5$
$d=12-10.5$.........(from 3)
$d=1.5$
$n=\frac{6}{1.5}=4$
Total number of terms are ${}^{\prime }2n^{\prime }$ i.e. $8$