Question

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

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

• A. 1
• B. 5
• C. 9
• D. 0

Answer 1

The correct option is D 0

$a_1=15a_3=2a_2-a_1{a}_4=2a_3-a_2=4a_2-2a_1-a_2=3a_2-2a_1{a}_5=2a_4-a_3=6a_2-4a_1-2a_2+a_1=4a_2-3a_1{a}_6=5a_2-6a_1-3a_1+2a_1=5a_2-4a_1{a}_7=2a_6-a_5=10a_2-8a_1-4a_2+3a_1=6a_2-5a_1{a}_k=(k-1)a_2-(k-2)a_{1}k\ge 3\frac{{a_1}^2+{a_2}^2+{a_3}^2+....+{a_{11}}^2}{11}=90$

$={a_1}^2+{a_2}^2+\stackrel{11}{\sum _{k=3}}[{(k-1)}^2{a_2}^2+{(k-2)}^2{a_1}^2-2(k-1\left)\right(k-2\left)a_1{a}_2\right]=990={a_1}^2+{a_2}^2+{a_2}^2\stackrel{11}{\sum _{k=3}}(k^2+1-2k)+\stackrel{11}{\sum _{k=3}}(k^2+4-4k){a_1}^2-2a_1{a}_2\stackrel{11}{\sum _{k=3}}(k^2-3k+2)=990$

$={a_1}^2+{a_2}^2+\stackrel{11}{\sum _{k=3}}({a_2}^2+{a_1}^2-2a_1{a}_2)k^2+\stackrel{11}{\sum _{k=3}}k(-2{a_2}^2-4{a_1}^2+6a_1{a}_2)+({a_2}^2+4{a_1}^2-4a_1{a}_2)\stackrel{11}{\sum _{k=3}}\left(1\right)=990={a_1}^2+{a_2}^2+({a_2}^2+{a_1}^2-2a_1{a}_2){\left[\frac{k(k+1)(2k+1)}{6}\right]}_2^{11}+(-2{a_2}^2-4{a_1}^2+6a_1{a}_2){\left[\frac{k(k+1)}{2}\right]}_2^{11}+({a_2}^2+4{a_1}^2-4a_1{a}_2){\left[k\right]}_2^{11}=990$

$={a_1}^2+{a_2}^2+({a_2}^2+{a_1}^2-2a_1{a}_2)\left[\frac{11\left(12\right)\left(23\right)-2\left(3\right)\left(5\right)}{6}\right]+(-2{a_2}^2-4{a_1}^2+6a_1{a}_2)\left[\frac{11\left(12\right)-2\left(3\right)}{2}\right]+({a_2}^2+4{a_1}^2-4a_1{a}_2)[11-2]=990$

$={a_1}^2+{a_2}^2+({a_2}^2+{a_1}^2-2a_1{a}_2)\left(501\right)+(-2{a_2}^2-4{a_1}^2+6a_1{a}_2)63+({a_2}^2+4{a_1}^2-4a_1{a}_2)9=990=(1+501+9-126){a_2}^2+{a_1}^2(1+501-252+36)+2a_1{a}_2(-501+189-18)=990(\because a_1=15)=385{a_2}^2+64350-9900a_2=990=385{a_2}^2-9900a_2+63360=0a_2=\frac{+9900\pm \sqrt{{\left(9900\right)}^2-4\times 385\times 63360}}{2\times 385}{a}_2=12\frac{(a_1+a_2+a_3+....+a_{11})}{11}=\frac{[a_1+a_2+\stackrel{11}{\sum _{k=3}}(k-1)a_2-(k-2\left)a_1\right]}{11}=\frac{[a_1+a_2+(a_2-a_1)\stackrel{11}{\sum _{k=3}}k+(2a_1-a_2\left)\stackrel{11}{\sum _{k=3}}\right(1\left)\right]}{11}=\frac{15+12+(-3){\left[\frac{k(k+1)}{2}\right]}_2^{11}+(2\times 15-12){\left[k\right]}_2^{11}}{11}=\frac{(27-3\times 63+18\times 9)}{11}=0$