Question

If $a_1,a_2,a_3,\dots$ are terms of AP such that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$, then the sum of first $24$ terms is

• A. $9\times {10}^2$
• B. $9\times {10}^3$
• C. $10\times 9^2$
• D. $10\times 9^3$

Answer 1

The correct option is A $9\times {10}^2$

We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow a_1+a_{24}=a_6+a_{20}=a_{10}+a_{15}$
$\because a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\therefore 3(a_1+a_{24})=225\Rightarrow a_1+a_{24}=75$
$\therefore S_{24}=\frac{24}{2}(a_1+a_{24})$ $[\because S_n=\frac{n}{2}(a_1+a_n)]$
$=12\left(75\right)=900=9\times {10}^2$