The sum of p terms of an A.P. is 3p^2 + 4p. F | vedclass

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(B) Given the sum of $p$ terms is $S_p = 3p^2 + 4p$.To find the $n^{th}$ term $(a_n)$,we use the relation $a_n = S_n - S_{n-1}$.First,substitute $p =

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$6n + 1$

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(B) Given the sum of $p$ terms is $S_p = 3p^2 + 4p$.To find the $n^{th}$ term $(a_n)$,we use the relation $a_n = S_n - S_{n-1}$.First,substitute $p = n$ to get $S_n = 3n^2 + 4n$.Next,substitute $p = n-1$ to get $S_{n-1} = 3(n-1)^2 + 4(n-1)$.Expanding $S_{n-1}$: $S_{n-1} = 3(n^2 - 2n + 1) + 4n - 4 = 3n^2 - 6n + 3 + 4n - 4 = 3n^2 - 2n - 1$.Now,calculate $a_n = S_n - S_{n-1} = (3n^2 + 4n) - (3n^2 - 2n - 1)$.$a_n = 3n^2 + 4n - 3n^2 + 2n + 1 = 6n + 1$.

The sum of $p$ terms of an $A.P.$ is $3p^2 + 4p$. Find the $n^{th}$ term.

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