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

Correct answer: (D) -72/5. Let a_1 be the first term and d be the common difference of A.P. ∴S_11=0 ⇒11/22a_1+10d=0 ⇒ a_1+5d=0 Now a_1+a_3+a_5+….+a_23=12/22a_1+12-12d =12a_1+11d =12a_1-11a_1/5 =-72a_1/5

Q40 / 75 · JEE Main 2020 (02 Sep Shift 2) · Mathematics

If the sum of first $11$ terms of an A.P. , $a_{1}, a_{2}, a_{3} \dots \dots$ is $0 (a_{1} \neq 0)$ then the sum of the A.P $a_{1}, a_{3}, a_{5}, \dots . . a_{23}$ is $k a_{1}$ where $k$ is equal to

A
$- \frac{121}{10}$
B
$\frac{121}{10}$
C
$\frac{72}{5}$
Correct answer
D
$- \frac{72}{5}$

Solution

Let $a_{1}$ be the first term and $d$ be the common difference of $A . P .$

$∴ S_{11} = 0 \Rightarrow \frac{11}{2} (2 a_{1} + 10 d) = 0$

$\Rightarrow a_{1} + 5 d = 0$

Now $a_{1} + a_{3} + a_{5} + \dots . + a_{23} = \frac{12}{2} (2 a_{1} + (12 - 1) 2 d)$ $= 12 (a_{1} + 11 d)$
$= 12 (a_{1} - \frac{11 a_{1}}{5})$
$= - \frac{72 a_{1}}{5}$

Hence, the correct answer is option: 'D'.

If the sum of first 11 terms of an A.P. ,a1,a2,a3…… is 0a1≠0 then the sum of the A.P a1,a3,a5,…..a23 is ka1 where k is equal to

Solution

Relevant Questions

If the sum of first $11$ terms of an A.P. , $a_{1}, a_{2}, a_{3} \dots \dots$ is $0 (a_{1} \neq 0)$ then the sum of the A.P $a_{1}, a_{3}, a_{5}, \dots . . a_{23}$ is $k a_{1}$ where $k$ is equal to