JEE MAIN - Mathematics (2020 - 2nd September Evening Slot - No. 20)
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 :
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 :
$${{121} \over {10}}$$
-$${{121} \over {10}}$$
$${{72} \over 5}$$
-$${{72} \over 5}$$
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 = $$ - {{72} \over 2}$$
$$ \because $$ 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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