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 :
Solution
Let common difference be d.
<br><br>$\because$ a<sub>1</sub>
+ a<sub>2</sub>
+ a<sub>3</sub>
+ ... + a<sub>11</sub> = 0
<br><br>$\therefore$ ${{11} \over 2}\left[ {2{a_1} + 10d} \right]$ = 0
<br><br>$\Rightarrow$ a<sub>1</sub>
+ 5d = 0
<br><br>$\Rightarrow$ d = ${ - {{{a_1}} \over 5}}$ .....(1)
<br><br>Now a<sub>1</sub>
+ a<sub>3</sub>
+ a<sub>5</sub>
+ ... + a<sub>23</sub>
<br><br>= (a<sub>1</sub> + a<sub>23</sub>) $\times$ ${{12} \over 2}$
<br><br>= (a<sub>1</sub> + a<sub>1</sub> + 22d) × 6
<br><br>= $\left[ {2{a_1} + 22\left( { - {{{a_1}} \over 5}} \right)} \right]$ $\times$ 6
<br><br>= $- {{72} \over 2}{a_1}$
<br><br>$\therefore$ k = $- {{72} \over 2}$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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