Question

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

• A. 909
• B. 75
• C. 750
• D. 900

Answer 1

Answer: (D)

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\Rightarrow 3\left(a_1+a_{24}\right)=225$
$(\because$ sum of terms equidistant from beginning and end are equal)
$\Rightarrow a_1+a_{24}=75$
Now $a_1+a_2+\dots \dots ..+a_{23}+a_{24}$
$=\frac{24}{2}\left[a_1+a_{24}\right]=12\times 75=900$