Find the sum of the two middle most terms of | vedclass

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(D) Here,the first term $(a) = -\frac{4}{3}$,common difference $(d) = -1 - (-\frac{4}{3}) = -1 + \frac{4}{3} = \frac{1}{3}$,and the last term $(l) = 4

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Here,the first term $(a) = -\frac{4}{3}$,common difference $(d) = -1 - (-\frac{4}{3}) = -1 + \frac{4}{3} = \frac{1}{3}$,and the last term $(l) = 4 \frac{1}{3} = \frac{13}{3}$.Since the $n^{th}$ term of an $AP$ is given by $l = a_n = a + (n - 1)d$,we have:$\frac{13}{3} = -\frac{4}{3} + (n - 1) \frac{1}{3}$Multiplying by $3$ on both sides:$13 = -4 + (n - 1)$$13 + 4 = n - 1$$n - 1 = 17$$n = 18$ (which is an even number).Since $n$ is even,the two middle most terms are the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ terms,i.e.,the $9^{th}$ and $10^{th}$ terms.$a_9 = a + 8d = -\frac{4}{3} + 8(\frac{1}{3}) = \frac{-4 + 8}{3} = \frac{4}{3}$.$a_{10} = a + 9d = -\frac{4}{3} + 9(\frac{1}{3}) = \frac{-4 + 9}{3} = \frac{5}{3}$.Sum of the two middle most terms $= a_9 + a_{10} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3$.

Find the sum of the two middle most terms of the $AP: -\frac{4}{3}, -1, -\frac{2}{3}, \ldots, 4 \frac{1}{3}$.

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