Let a1- a2- a3- - a11 be real numbers satisfying a1-15-27-2a2-0 and ak-2ak-1-ak-2 for k-3-4- - 11- If a12-a22-a112-11-90- then the value of a1-a2-a11-11 is equal to

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

Let a1, a2,...

Question

Let $a_{1}, a_{2}, a_{3}, \dots, a_{11}$ be real numbers satisfying $a_{1} = 15, 27 - 2 a_{2} > 0$ and $a_{k} = 2 a_{k - 1} - a_{k - 2}$ for $k = 3, 4, \dots, 11.$ If $\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{11}^{2}}{11} = 90$ , then the value of $\frac{a_{1} + a_{2} + \dots + a_{11}}{11}$ is equal to

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Solution

The correct option is D $0$
We have: $a_{k} + a_{k + 2} = 2 a_{k - 1}$
Thus, the terms are in AP.
Hence, sum of squares of the terms in AP is:
$a^{2} + {(a + d)}^{2} + . . . + {(a + 10 d)}^{2} = 11 a^{2} + 110 a d + 385 d^{2} = 990 => a^{2} + 10 a d + 35 d^{2} = 90 => 35 d^{2} + 150 d + 225 - 90 = 0 => 35 d^{2} + 150 d + 135 = 0 => 7 d^{2} + 30 d + 27 = 0 => (7 d + 9) (d + 3) = 0 => d = - 3, - \frac{9}{7}$
Since $a_{2} < 13.5$ , $d = - 3$
Thus, the required answer i.e. the average of $11$ terms of an AP = $a_{6} = 15 + (6 - 1) . (- 3) = 0$
Hence, (a) is correct.

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Similar questions

Q. Let

a_{1}, a_{2}, a_{3}, . . . ., a_{11}

be real numbers satisfying

a_{1} = 15, 27 - 2 a_{2} > 0

and

a_{k} = 2 a_{k - 1} - a_{k - 2}

for

k = 3, 4, . . . . . . ., 11

\frac{a_{1}^{2} + a_{2}^{2} + . . . + a_{11}^{2}}{11} = 90

, then the value of

\frac{a_{1} + a_{2} + . . . + a_{11}}{11}

is equal to

Let $a_{1}, a_{2}, a_{3}, . . . ., a_{11}$ be real numbers satisfying $a_{1} = 15, 27 - 2 a_{2} > 0$ and $a_{k} = 2 a_{k - 1} - a_{k - 2}$ for k = 3, 4, ….11. If $\frac{a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + . . . . a_{11}^{2}}{11} = 90$ , then $\frac{a_{1} + a_{2} + a_{3} + . . . . . a_{11}}{11}$ is equal to