Question

There are $2n$ terms in an A.P., whose first term is $a$ and common difference is $d$. The sum of the odd terms is $24$ and the sum of the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$, then which of the following is (are) CORRECT?

• A. $a=\frac{3}{2}$
• B. $d=\frac{5}{2}$
• C. $n=4$
• D. $n=8$

Answer 1

Answer: (A), (C)

The correct options are
A $a=\frac{3}{2}$
C $n=4$

$a,a+d,a+2d,a+3d,\dots$ are in A.P. ($2n$ terms)
Sum of the odd terms $=\frac{n}{2}[2a+(n-1\left)2d\right]=24$
$\Rightarrow n[a+(n-1\left)d\right]=24\cdots \left(1\right)$

Sum of the even terms$=\frac{n}{2}\left[2\right(a+d)+(n-1\left)2d\right]=30$
$\Rightarrow n[a+nd]=30\cdots \left(2\right)$

$a_{2n}-a=10\frac{1}{2}$
$\Rightarrow a+(2n-1)d-a=\frac{21}{2}$
$\Rightarrow (2n-1)d=\frac{21}{2}\cdots \left(3\right)$
From $\left(2\right)-\left(1\right)$, we get $nd=6$
From $\left(3\right)$,
$12-d=\frac{21}{2}\Rightarrow d=\frac{3}{2}$
$\therefore n\left(\frac{3}{2}\right)=6\Rightarrow n=4$
From $\left(2\right)$,
$4(a+6)=30\Rightarrow a=\frac{3}{2}$