Question

If $a_1,a_2,a_3,............$ is an A.P. such that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225,$ then $a_1+a_2+a_3+.....+a_{23}+a_{24}$ is equal to

• A. 909
• B. 75
• C. 750
• D. 900

Answer 1

Answer: (D)

We have
$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\therefore \left(a_1+a_{24}\right)+\left(a_5+a_{20}\right)+\left(a_{10}+a_{15}\right)=225$
$\therefore \left(a+a+23d\right)+\left(a+4d+a+19d\right)$
$+\left(a+9d+a+14d\right)=225$
i.e., $3\left(2a+23d\right)=225$
$\Rightarrow 2a+23d=75$
Again $a_1+a_2+.....+a_{24}$
$=\frac{24}{2}\left(a_1+a_{24}\right)=12\left(a+a+23d\right)$
$=12\left(2a+23d\right)$
$=12\left(75\right)$
$=900$