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 $T_{n} = 10 - 6n$.To find the first term $a$,substitute $n = 1$:$a = T_{1} = 10 - 6(1) = 4$.To find the s

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$S_{n} = -3n^{2} + 7n$

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(A) Given the $n^{th}$ term of the $A.P.$ is $T_{n} = 10 - 6n$.To find the first term $a$,substitute $n = 1$:$a = T_{1} = 10 - 6(1) = 4$.To find the second term $T_{2}$,substitute $n = 2$:$T_{2} = 10 - 6(2) = 10 - 12 = -2$.The common difference $d = T_{2} - T_{1} = -2 - 4 = -6$.The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_{n} = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values of $a$ and $d$:$S_{n} = \frac{n}{2} [2(4) + (n - 1)(-6)]$$S_{n} = \frac{n}{2} [8 - 6n + 6]$$S_{n} = \frac{n}{2} [14 - 6n]$$S_{n} = n(7 - 3n) = 7n - 3n^{2}$.Thus,$S_{n} = -3n^{2} + 7n$.

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 10 - 6n$. Find the sum of the first $n$ terms of the $A.P.$

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