Love Algebra (Simple "FUN" Problem)

Level 2

Let $a_{1},a_{2},a_{3}.....a_{11}$ be real numbers

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

$a_{k}=2a_{k-1}-a_{k-2}$ for $k=3,4,5.....11.$

If $\frac{a^{2}_{1}+a^{2}_{2}+a^{2}_3.....+a^{2}_{11}}{11}=90$

then the value of

$\frac{a_{1}+a_{2}+a_{3}....a_{11}}{11}$ is equal to

The answer is 0.

1 solution

Rajdeep Brahma
Apr 15, 2017

We can write the recursion as follows:ak-a(k-1)=a(k-1)-a(k-2)=d
Hence this is an AP.
a1=15,a2=15+d,...,a11=15+10d
15^2+(15+d)^2+...+(15+10d)^2=990
(15^2X15)+2X15XdX(1+2+...+10)+(1^2+2^2+...+10^2)=990
Further simplifying we get,
7d^2+30d+27=0
d=(-3)or(- $\frac{9}{7}$ )
Since 13.5>a2,so d=(-3)
$\frac{15+(15+d)+...+(15+10d)}{11}$ =15+5d=15-15=0(ANSWER)

Love Algebra (Simple "FUN" Problem)

The answer is 0.

1 solution

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