If the sum of first 11 terms of an A.P.,n a_1,a_2,a_3, ... .. is 0( a_1 ≠ 0 ),n then the sum of the A.P., a_1,a_3,a_5, ... ,a_23 isn ka_1, where k is equal to

(c) : Given,n a_1 + a_2 + a_3 ... ... ... ... + a_11 = 0n nnLet d be the common difference of given A.P. nThen,nS_11 = (11 / 2)( 2a_1 + 10d ) = 0 => 10d = - 2a_1 => d = frac- a_15 nNow; givenna_1 + a_3 + a_5 + ... ... ... ... + a_23 = ka_1 n => (12 / 2)( 2a_1 + 11(2d) ) = ka_1 lbrack Because a_2 - a_1 = d Therefore a_3 - a_1 = 2d rbracknn => 6( 2a_1 - frac22a_15 ) = ka_1 => k = 6( (10 - 22 / 5) ) = (- 72 / 5)

If the sum of first 11 terms of an A.P., a_{1},a_{2},... | Sequence & Series (Mathematics)

a_{1} + a_{2} + a_{3}\ldots\ldots\ldots\ldots + a_{11} = 0

\n\end{enumerate}\n\nLet

d

be the common difference of given A.P.\\\nThen,\n

S_{11} = \frac{11}{2}\left( 2a_{1} + 10d \right) = 0 \Rightarrow 10d = - 2a_{1} \Rightarrow d = \frac{- a_{1}}{5}

\\\nNow; given\n

a_{1} + a_{3} + a_{5} + \ldots\ldots\ldots\ldots + a_{23} = ka_{1}

\\\n

\Rightarrow \frac{12}{2}\left( 2a_{1} + 11(2d) \right) = ka_{1}\ \left\lbrack \because a_{2} - a_{1} = d\ \therefore a_{3} - a_{1} = 2d \right\rbrack\n

\Rightarrow 6\left( 2a_{1} - \frac{22a_{1}}{5} \right) = ka_{1} \Rightarrow k = 6\left( \frac{10 - 22}{5} \right) = \frac{- 72}{5}

Explanation

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