Question

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

Answer 1

Sum of all the terms is $24+30=54$
Sum $=\frac{n(2a+(n-1\left)d\right)}{2}$ where a is the first term, $d$ is the common difference, $n$ is the number of terms.
Also, $(n-1)d=10.5$
So, $54=\frac{n(2a+10.5)}{2}$
Also, $30=\frac{n(2a+(n-1)d+d)}{4}$
i.e. $30=\frac{n(2a+10.5+d)}{4}$
$\therefore$ we get $108=2an+10.5n$ and $120=2an+10.5n+dn$
Subtracting the two, we have $12=dn$
But $nd-d=10.5$, implying $d=1.5$ & $n=8$
Also, $108=2an+10.5n=16a+84$
$\therefore a=1.5$
Thus, second term = $a+d=3$
Hence $3+8$ equals $11.$