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?
​

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.

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

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

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

Which is the required Answer.

Answer 2

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?

To Find:-

• Find the value of k.

Solution:-

• Let d be the common difference

Formula to be Used:-

• $\tt\implies \: S_n = \dfrac { n } { 2 } ( 2a + ( n - 1 )d )$

$\tt\implies \: \dfrac { 11 } { 2 } ( 2( a_1 + ( 11 - 1 )d )$

$\tt\implies \: a_1 + 5d . . . . ( 1 )$

According to the Question:-

$\tt\implies \: a_1 + + a_3 + a_5 + . . . . + a_23 = ka_1$

$\tt\implies \: \cancel\dfrac { 12 } { 2 } [ 2a_1 + ( 12 - 1 )2d ] = ka_1$

$\tt\implies \: 6( \dfrac { 10a_1 - 22a_1 } { 5 } ) = 5a_1$

$\tt\implies \: k = \dfrac { - 72 } { 5 }$

Hence ,

• The value of k = $\tt \: \dfrac { - 72 } { 5 }$