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 $10.5$, then the number of terms in the series is

• A. $8$
• B. $4$
• C. $6$`
• D. $10$

Answer 1

The correct option is A $8$

Let the total number of terms be '2n', first term be 'a' and common difference be 'd'

$a+(2n-1)d=l$

and $l-a=\mathrm{10.5.......................}\left(1\right)$

if we take only odd terms, it will also be a A.P of n terms with common difference '2d'

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

$24=n\{a+(n-1\left)d\right\}................\left(2\right)$

Similarly,

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

$30=\frac{n}{2}\{2a+2nd\}$

$30=n\{a+nd\}........................\left(3\right)$

Putting value of $n\{a+nd\}$from eq (3) to eq (2)

$24=30-nd$

$nd=6$

putting value of $nd$ in eq(1)

$12-d=10.5$

$d=1.5$

So,

$n=\frac{6}{1.5}=4$

So, number of terms will be $2n=8$