Question

If the sum of first 11 terms of an AP.,
a1,a2 ,... is 0 (a1 not equals to 0), then the sum of the AP.,
a1, a3, a5 ,..., a23 is ka1, where k is equal to?
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Answer 1

ANSWER:-)

$\dfrac{ - 72}{5}$

EXPLANATION :-)

Let d be the common difference of the given AP

According to the question

Sum first 11 terms of the AP is 0

i.e.

$S_{11} = 0 \\ \\ \implies \frac{11}{2} (2a_1 + (11 - 1)d) = 0 \\ \\ a_1 + 5d = 0 \: \: \: \: \: \: \: \: \: ..(i)$

Now Given that sum of the AP

i.e.

$a_1 + a_3 + a_5 + .. + a_{23} =ka_1 \\ \\ \frac{12}{2} (2a_1 + (12 - 1)(2d)) = ka_1$

$6(2a_1 + 22 (\frac{ -a_1}{5} )) = 5a_1 \: \: \: ....(using \: (i))$

$6( \frac{10a_1 - 22a_1}{5} ) = 5a_1 \\ \\ \purple{ \boxed{ \boxed{k = \frac{ - 72}{5} }}}$

Answer 2

ANSWER:-)

$\dfrac{ - 72}{5}$

EXPLANATION :-)

Let d be the common difference of the given AP

According to the question

Sum first 11 terms of the AP is 0

i.e.

$S_{11} = 0 \\ \\ \implies \frac{11}{2} (2a_1 + (11 - 1)d) = 0 \\ \\ a_1 + 5d = 0 \: \: \: \: \: \: \: \: \: ..(i)$

Now Given that sum of the AP

i.e.

$a_1 + a_3 + a_5 + .. + a_{23} =ka_1 \\ \\ \frac{12}{2} (2a_1 + (12 - 1)(2d)) = ka_1$

$6(2a_1 + 22 (\frac{ -a_1}{5} )) = 5a_1 \: \: \: ....(using \: (i))$

$6( \frac{10a_1 - 22a_1}{5} ) = 5a_1 \\ \\ \pink{ \boxed{ \boxed{k = \frac{ - 72}{5} }}}$