Consider an A.P.a1,a2,a3,… such that a3+a5+a8=11 and a4+a2=−2 then the value of a1+a6+a7 is

A
-8
B
5
C
7
D
9

Solution:

Given that $a_{7} + a_{5} + a_{8} = 11$
$\Rightarrow a + 2 d + a + 4 d + a + 7 d = 11$
or $3 a + 13 d = 11 (1)$
Given,
$a_{4} + a_{2} = - 2$
$\Rightarrow a + 3 d + a + d = - 2$
or $a = - 1 - 2 d (2)$
Putting value of a from (2) in (1), we get
$3 (- 1 - 2 d) + 13 d = 11 \Rightarrow 7 d = 14 \Rightarrow d = 2 and a = - 5$
$\Rightarrow a_{1} + a_{6} + a_{7} = 7$

Consider an A.P. $a_{1}, a_{2}, a_{3}, \dots$ such that $a_{3} + a_{5} + a_{8} = 11$ and $a_{4} + a_{2} = - 2$ then the value of $a_{1} + a_{6} + a_{7}$ is

Solution:

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