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$