Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{11}$ be real numbers satisfying \[ \mathrm{a}_{1}=15, \quad 27-2 \mathrm{a}_{2}>0 \text { and } \mathrm{a}_{\mathrm{k}}=2 \mathrm{a}_{\mathrm{k}-1}-\mathrm{a}_{\mathrm{k}-2} \text { for } \mathrm{k}=3,4, \ldots, 11 . \] If $\frac{\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots+\mathrm{a}_{11}^{2}}{11}=90$, then the value of $\frac{\mathrm{a}_{1}+\mathrm{a}_{2}+\ldots+\mathrm{a}_{11}}{11}$ is equal to

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Sequences and Series

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.

\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

Answer: 0

The recurrence relation

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

implies that the sequence is an arithmetic progression (AP).

Let the common difference be

d

. Then

a_{n} = a_{1} + (n - 1) d = 15 + (n - 1) d

.

Given:

\frac{\sum _{i = 1}^{11} a _{i}^{2}}{11} = 90

For an AP:

i = 1 \sum 11 a_{i}^{2} = 11 a_{1}^{2} + 2 a_{1} d i = 0 \sum 10 i + d^{2} i = 0 \sum 10 i^{2}

= 11 \cdot 225 + 2 \cdot 15 \cdot d \cdot 55 + d^{2} \cdot 385

= 2475 + 1650 d + 385 d^{2}

Setting this equal to

11 \cdot 90 = 990

2475 + 1650 d + 385 d^{2} = 990

385 d^{2} + 1650 d + 1485 = 0

77 d^{2} + 330 d + 297 = 0

7 d^{2} + 30 d + 27 = 0

Using the quadratic formula:

d = \frac{- 30 \pm 900 - 756}{14} = \frac{- 30 \pm 12}{14}

d = - \frac{9}{7} or d = - 3

Given

27 - 2 a_{2} > 0

, we have

a_{2} < 13.5

. Since

a_{2} = 15 + d

d = - 3

.

The mean is:

\frac{\sum _{i = 1}^{11} a _{i}}{11} = \frac{11 a _{1} + d \cdot 55}{11} = a_{1} + 5 d = 15 + 5 (- 3) = 0

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Integer answer