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(C) For the given $A.P.$,the first term $a = 7$ and the $10^{th}$ term $T_{10} = 61$.The formula for the $n^{th}$ term of an $A.P.$ is $T_{n} = a + (n

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$151$

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(C) For the given $A.P.$,the first term $a = 7$ and the $10^{th}$ term $T_{10} = 61$.The formula for the $n^{th}$ term of an $A.P.$ is $T_{n} = a + (n - 1)d$.For the $10^{th}$ term: $T_{10} = a + 9d$.Substituting the values: $61 = 7 + 9d$.$54 = 9d$,which gives $d = 6$.Now,to find the $25^{th}$ term: $T_{25} = a + 24d$.$T_{25} = 7 + 24(6) = 7 + 144 = 151$.Thus,the common difference is $6$ and the $25^{th}$ term is $151$.

For a given $A.P.$,the first term is $7$ and the $10^{th}$ term is $61$. Find the common difference of the $A.P.$ and its $25^{th}$ term.

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