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If the sum of first 11 terms of an A.P., a1, a2, a3, .... is 0 (a $$ \ne $$ 0), then the sum of the A.P., a1 , a3 , a5 ,....., a23 is ka1 , where k is equal to :

JEE · Math · previous-year question

  1. A.$${{121} \over {10}}$$
  2. B.-$${{121} \over {10}}$$
  3. C.$${{72} \over 5}$$
  4. D.-$${{72} \over 5}$$correct

Answer

D. -$${{72} \over 5}$$

Explanation

Let common difference be d. $$ \because $$ a1 + a2 + a3 + ... + a11 = 0 $$ \therefore $$ $${{11} \over 2}\left[ {2{a_1} + 10d} \right]$$ = 0 $$ \Rightarrow $$ a1 + 5d = 0 $$ \Rightarrow $$ d = $${ - {{{a_1}} \over 5}}$$ .....(1) Now a1 + a3 + a5 + ... + a23 = (a1 + a23) $$ \times $$ $${{12} \over 2}$$ = (a1 + a1 + 22d) × 6 = $$\left[ {2{a_1} + 22\left( { - {{{a_1}} \over 5}} \right)} \right]$$ $$ \times $$ 6 = $$ - {{72} \over 2}{a_1}$$ $$ \therefore $$ k = $$ - {{72} \over 2}$$

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