Question

Let a1, a2, a3, …… , a11 be real numbers satisfying a1 = 15, 27-2 a2> 0 and ak = 2ak-1 - ak-2 for k = 3, 4, …. , 11. If (a12 + a22 + a32 + …… + a112)/11 = 90, then find the value of (a1 + a2 + a3 + …… + a11)/11

Answer 1

Answer:

Given that ak = 2ak-1 - ak-2

This relation clearly implies that a1, a2, a3, …… , a11 are in A.P.

Hence, (a1^2 + a2^2 + a3^2 + …… + a11^2)/11 = [11a^2 + 35.11d^2 +10ad]/11 = 90.

Hence, we obtain, 225 + 35.d^2 +150d = 90

So, 35.d^2 +150d + 135 = 0

This gives d = -3, -9/7

But, it is given that a2< 27/2 and hence, d = -3, d ≠ -9/7.

So, (a1 + a2 + a3 + …… + a11)/11 = 11/2 [30-10.3] = 0.

Hope it will help you

Answer 2

Answer:

our hero has already answered please refer it

Step-by-step explanation:

Hi friend im new to brainly

please thank me thank me.....