The n^{th} term of an A.P. is given by T_{n} | vedclass

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(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 4n + 7$.To find the first term $a$,substitute $n = 1$:$a = T_{1} = 4(1) + 7 = 11$.To find the se

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$2n^{2} + 9n$

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(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 4n + 7$.To find the first term $a$,substitute $n = 1$:$a = T_{1} = 4(1) + 7 = 11$.To find the second term,substitute $n = 2$:$T_{2} = 4(2) + 7 = 15$.The common difference $d$ is given by $d = T_{2} - T_{1} = 15 - 11 = 4$.The sum of $n$ terms of an $A.P.$ is given by the formula $S_{n} = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values $a = 11$ and $d = 4$:$S_{n} = \frac{n}{2}[2(11) + (n - 1)4]$$S_{n} = \frac{n}{2}[22 + 4n - 4]$$S_{n} = \frac{n}{2}[4n + 18]$$S_{n} = n[2n + 9]$$S_{n} = 2n^{2} + 9n$.Thus,the sum of $n$ terms is $2n^{2} + 9n$.

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 4n + 7$. Find the sum of $n$ terms of this $A.P.$

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