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-a11211-90- then the value of a-1-a2-a1111 is equal to

Formula of ∑_r=1^nr=n(n+1)2 and ∑_r=1^nr^2=n(n+1)(2n+1)6 are used. ...

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Byju's Answer

Standard VIII

Mathematics

Division of a Polynomial by a Monomial

Let a1,a2,a3,...

Question

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 $\frac{a_{1}^{2} + a_{2}^{2} + . . = a_{11}^{2}}{11} = 90$ , then the value of $\frac{a - 1 + a_{2} + . . + a_{11}}{11}$ is equal to

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Solution

Formula of $\sum_{r = 1}^{n} r = \frac{n (n + 1)}{2}$ and $\sum_{r = 1}^{n} r^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ are used.

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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

Q.
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

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