If the middle term of the AP is 300,then the | vedclass

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(A) Given,number of terms,$n = 51$. Since $n$ is odd,the middle term is the $\left(\frac{n+1}{2}ight)$-th term. $	ext{Middle term} = \left(\frac{51

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$15300$

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(A) Given,number of terms,$n = 51$. Since $n$ is odd,the middle term is the $\left(\frac{n+1}{2}ight)$-th term. $	ext{Middle term} = \left(\frac{51+1}{2}ight) = 26	ext{-th term}$. Thus,$T_{26} = a + 25d = 300$. The sum of the first $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$,where $l$ is the last term. Here,$l = T_{51} = a + 50d$. $S_{51} = \frac{51}{2}(a + a + 50d) = \frac{51}{2}(2a + 50d) = 51(a + 25d)$. Substituting $a + 25d = 300$,we get: $S_{51} = 51 	imes 300 = 15300$.

If the middle term of the $AP$ is $300$,then the sum of its first $51$ terms is

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