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Q.
If the first term of an $A.P$. is $3 $ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms, then the common difference of this $A.P$. is:
Sum of 1 st 25 terms $=$ sum of its next 15 terms
$\Rightarrow \left( T _{1}+\ldots \ldots+ T _{25}\right) =\left( T _{26}+\ldots . .+ T _{40}\right)$
$\Rightarrow \left( T _{1}+\ldots . .+ T _{40}\right) =2\left( T _{1}+\ldots \ldots+ T _{25}\right)$
$\Rightarrow \frac{40}{2}[2 \times 3+(39 d )]=2 \times \frac{25}{2}[2 \times 2+24 d ]$
$\Rightarrow d =\frac{1}{6}$