IIT-JEE 2010 · Paper 2 · Mathematics Q29

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MathematicsInteger+3Question 29 of 57

Let $a_1,a_2,a_3,\ldots,a_{11}$ be real numbers satisfying $a_1=15,\quad 27-2a_2>0\ \text{ and }\ a_k=2a_{k-1}-a_{k-2}\ \text{ for } k=3,4,\ldots,11.$ If $\dfrac{a_1^2+a_2^2+\cdots+a_{11}^2}{11}=90$ , then the value of $\dfrac{a_1+a_2+\cdots+a_{11}}{11}$ is equal to

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AP. $a_k=2a_{k-1}-a_{k-2}$ means an arithmetic progression: $a_k=15+(k-1)d$ .

Mean of squares. With $j=0,\ldots,10$ : $\sum j=55$ , $\sum j^2=385$ :
$\frac1{11}\sum(15+jd)^2=225+150d+35d^2=90\implies7d^2+30d+27=0\implies d=-3\text{ or }-\frac97.$

Condition $27-2a_2>0$ : $a_2=15+d<13.5$ , so $d<-1.5$ and $d=-3$ .

Mean. $\frac1{11}\sum a_k=15+5d=0$ .

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Topics:Arithmetic progression

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IIT-JEE 2010 · Paper 2, Mathematics question 29