Let a_1, a_2, ......., a_{30} be an A.P.,S = | vedclass

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(A) Given that $a_1, a_2, ......., a_{30}$ is an $A.P.$ with first term $a$ and common difference $D$.$S = a_1 + a_2 + a_3 + ....... + a_{30} = \frac{

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$52$

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(A) Given that $a_1, a_2, ......., a_{30}$ is an $A.P.$ with first term $a$ and common difference $D$.$S = a_1 + a_2 + a_3 + ....... + a_{30} = \frac{30}{2} [2a + 29D] = 15(2a + 29D) = 30a + 435D$.$T = a_1 + a_3 + a_5 + ....... + a_{29}$ is an $A.P.$ with $15$ terms,first term $a_1 = a$,and common difference $2D$.$T = \frac{15}{2} [2a + (15-1)(2D)] = \frac{15}{2} [2a + 28D] = 15(a + 14D) = 15a + 210D$.Now,$S - 2T = (30a + 435D) - 2(15a + 210D) = 30a + 435D - 30a - 420D = 15D$.Given $S - 2T = 75$,so $15D = 75$,which implies $D = 5$.Given $a_5 = 27$,we have $a + 4D = 27$.Substituting $D = 5$,$a + 4(5) = 27 \Rightarrow a + 20 = 27 \Rightarrow a = 7$.We need to find $a_{10} = a + 9D$.$a_{10} = 7 + 9(5) = 7 + 45 = 52$.

Let $a_1, a_2, ......., a_{30}$ be an $A.P.$,$S = \sum_{i=1}^{30} a_i$ and $T = \sum_{i=1}^{15} a_{2i-1}$. If $a_5 = 27$ and $S - 2T = 75$,then $a_{10}$ is equal to

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