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

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(A) Given $S = \sum_{i=1}^{30} {a_i}$ and $T = \sum_{i=1}^{15} {a_{2i-1}}$.Let the $A.P.$ be defined as ${a_i} = a + (i-1)d$.$S = {a_1} + {a_2} + {a_3

Vedclass · Accepted Answer

$52$

Vedclass · Answer

(A) Given $S = \sum_{i=1}^{30} {a_i}$ and $T = \sum_{i=1}^{15} {a_{2i-1}}$.Let the $A.P.$ be defined as ${a_i} = a + (i-1)d$.$S = {a_1} + {a_2} + {a_3} + \dots + {a_{30}}$$T = {a_1} + {a_3} + {a_5} + \dots + {a_{29}}$Then $2T = 2{a_1} + 2{a_3} + 2{a_5} + \dots + 2{a_{29}}$.$S - 2T = ({a_2} - {a_1}) + ({a_4} - {a_3}) + ({a_6} - {a_5}) + \dots + ({a_{30}} - {a_{29}})$.Since ${a_{2k}} - {a_{2k-1}} = d$,we have $S - 2T = 15d$.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 ${a_1}, {a_2}, \dots, {a_{30}}$ 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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