If the n^{th} term of an arithmetic progressi | vedclass

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(C) Given the $n^{th}$ term $T_n = \frac{2n + 1}{3}$.For $n = 1$,$a = T_1 = \frac{2(1) + 1}{3} = \frac{3}{3} = 1$.For $n = 2$,$T_2 = \frac{2(2) + 1}{3

Vedclass · Accepted Answer

$133$

Vedclass · Answer

(C) Given the $n^{th}$ term $T_n = \frac{2n + 1}{3}$.For $n = 1$,$a = T_1 = \frac{2(1) + 1}{3} = \frac{3}{3} = 1$.For $n = 2$,$T_2 = \frac{2(2) + 1}{3} = \frac{5}{3}$.The common difference $d = T_2 - T_1 = \frac{5}{3} - 1 = \frac{2}{3}$.The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n - 1)d]$.For $n = 19$,$S_{19} = \frac{19}{2} [2(1) + (19 - 1)(\frac{2}{3})]$.$S_{19} = \frac{19}{2} [2 + 18 	imes \frac{2}{3}] = \frac{19}{2} [2 + 12] = \frac{19}{2} 	imes 14 = 19 	imes 7 = 133$.

If the $n^{th}$ term of an arithmetic progression is $\frac{(2n + 1)}{3}$,what is the sum of its first $19$ terms?

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