Let {a₁},{a₂},.......,{a_{30}} be an A.P. , S = Σlimits

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8. Sequences and Series

hard

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

$52$

$57$

$47$

$42$

(JEE MAIN-2019)

Solution

$S = \sum\limits_{i – 1}^{30} {{a_i}} \,\,\,\,,T = \sum\limits_{i – 1}^{15} {{a_{2i – 1}}} \,\,\,\,\,,{a_5} = 27,S – 2T = 75$

Let ${a_i} = a + \left( {i – 1} \right)D$

$S = {a_1} + {a_2} + {a_3} + ……….. + {a_{30}}$

$T = {a_1} + {a_3} + {a_5} + ……….. + {a_{29}}$

$\therefore 2T = 2{a_1} + 2{a_3} + 2{a_5} + ……….. + 2{a_{29}}$

$S – 2T = \left( {{a_2} – {a_1}} \right) + \left( {{a_4} – {a_3}} \right) + \left( {{a_6} – {a_5}} \right) + …….. + \left( {{a_{30}} – {a_{29}}} \right) = 75$

$ = 15D$

But $S – 2T = 75 \Rightarrow 15D = 75 \Rightarrow D = 5$

Now ${a_5} = 27 \Rightarrow a + 4D = 27$

$\therefore a = 27 – 20 \Rightarrow a = 7$

${a_{10}} = a + 9D$

$ = 7 + 45 = 52$

Standard 11

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