Question

If $a_1,a_2,a_3,...$ 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

Answer: (A)

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.
$\Rightarrow a_1+a_{24}=a_5+a_{20}=a_{10}+a_{15}$
$\because a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\therefore 3\left(a_1+a_{24}\right)=225$
$\Rightarrow a_1+a_{24}=75$
$\therefore S_{24}=\frac{24}{2}\left(a_1+a_{24}\right)$
$\left[\because S_n=\frac{n}{2}\left(a_1+a_n\right)\right]$
$=12(75)=900=9\times 10^2$