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Q.
If $a_1, a_2, a_3, \ldots \ldots$. are in A.P. such that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$, then $a_1+a_2+a_3+\ldots .+a_{23}+a_{24}$ is equal to
Sequences and Series
Solution:
$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$
$\Rightarrow 3\left(a_1+a_{24}\right)=225$
$ (\because$ sum of terms equidistant from beginning and end are equal)
$\Rightarrow a_1+a_{24}=75$
Now $ a_1+a_2+\ldots \ldots . .+a_{23}+a_{24}$
$=\frac{24}{2}\left[a_1+a_{24}\right]=12 \times 75=900$