Question

consider an AP. a1,a2,a3.....such that a3+a5+a8=11and a4+a2=-2 then the value of a1+a6+a7 is

Answer 1

Hope it will help you..........

Answer 2

Answer: 7

Step-by-step explanation:

Given that $a_1,a_2,a_3........$ are in AP

such that

$a_3+a_5+a_8=11$ and $a_4+a_2=-2$

Now

$a_3+a_5+a_8=11\\\\\Rightarrow a+2d+a+4d+a+7d= 11\\\\\Rightarrow 3a+13d= 11 -----(i)$

Similarly

$a_4+a_2=-2\\\\\Rightarrow a+3d+a+d=-2\\\\\Rightarrow 2a+4d=-2\\\\\Rightarrow a+2d=-2 ------(ii)$

from (i) and (ii)

we get d= 2 and a= -5

$a_1+a_6+a_7= a+a+5d+a+6d= 3a+11d\\\\= 3\times (-5) + 11\times 2= -15+22= 7$

Hence, the required value is 7