Let a_1,\,a_2\,,\,a_3 ......, a_11 be real numbers satisfying a_1 = 15,27 - 2a_2 \gt 0\,\,and\,\,a_k = 2a_k - 1 - a_k - 2\,\,for\,k = 3,4,........11 . if \,\,\,a_1^2 + a_2^2 + .... + a_11^2 \over 11 = 90 , then the value of a_1 + a_2 + .... + a_11 \over 11 is equal to :

Sequences and Series PYQ — JEE Advanced 2010 · Shift 2 · Q26 with Solution

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Correct answer: 0

Interpret the recurrence

Given

\qquad 27-2a_2>0, \qquad a_k=2a_{k-1}-a_{k-2}\ \text{for } k=3,4,\dots,11.$$ The recurrence $$a_k- a_{k-1}=a_{k-1}-a_{k-2}$$ shows that the sequence is an **arithmetic progression**. So let the common difference be $d$. Then $$a_n=a_1+(n-1)d=15+(n-1)d.$$ In particular, $$a_2=15+d.$$ Also, $$27-2a_2>0 \Rightarrow 27-2(15+d)>0 \Rightarrow 27-30-2d>0 \Rightarrow -3-2d>0 \Rightarrow d< -\frac32.$$ --- 2. **Use the condition on the mean of squares** We are given $$\frac{a_1^2+a_2^2+\cdots+a_{11}^2}{11}=90.$$ Hence $$a_1^2+a_2^2+\cdots+a_{11}^2=990.$$ Now $$a_n=15+(n-1)d,$$ so for $r=0,1,2,\dots,10$, $$a_{r+1}=15+rd.$$ Thus $$\sum_{r=0}^{10}(15+rd)^2=990.$$ Expand: $$\sum_{r=0}^{10}(225+30rd+r^2d^2)=990.$$ So $$11\cdot 225+30d\sum_{r=0}^{10}r+d^2\sum_{r=0}^{10}r^2=990.$$ Using $$\sum_{r=0}^{10}r=55, \qquad \sum_{r=0}^{10}r^2=385,$$ we get $$2475+30d(55)+385d^2=990.$$ That is, $$2475+1650d+385d^2=990,$$ $$385d^2+1650d+1485=0.$$ Divide by $55$: $$7d^2+30d+27=0.$$ Factor: $$(7d+9)(d+3)=0.$$ So $$d=-\frac97 \quad \text{or} \quad d=-3.$$ From the condition $d< -\frac32$, only $$d=-3$$ is valid. --- 3. **Find the mean of the 11 terms** Since the sequence is an AP with first term $15$ and common difference $-3$, $$a_{11}=15+10(-3)=-15.$$ The average of an AP is $$\frac{a_1+a_{11}}{2}= rac{15+(-15)}{2}=0.$$ Therefore, $$\frac{a_1+a_2+\cdots+a_{11}}{11}=0.$$ --- 4. **Compare with stored answer** Derived answer: $0$. Stored correct answer: $0$. They agree.

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View written solutionFree

Correct answer: 0

Interpret the recurrence

Given

\qquad 27-2a_2>0, \qquad a_k=2a_{k-1}-a_{k-2}\ \text{for } k=3,4,\dots,11.$$ The recurrence $$a_k- a_{k-1}=a_{k-1}-a_{k-2}$$ shows that the sequence is an **arithmetic progression**. So let the common difference be $d$. Then $$a_n=a_1+(n-1)d=15+(n-1)d.$$ In particular, $$a_2=15+d.$$ Also, $$27-2a_2>0 \Rightarrow 27-2(15+d)>0 \Rightarrow 27-30-2d>0 \Rightarrow -3-2d>0 \Rightarrow d< -\frac32.$$ --- 2. **Use the condition on the mean of squares** We are given $$\frac{a_1^2+a_2^2+\cdots+a_{11}^2}{11}=90.$$ Hence $$a_1^2+a_2^2+\cdots+a_{11}^2=990.$$ Now $$a_n=15+(n-1)d,$$ so for $r=0,1,2,\dots,10$, $$a_{r+1}=15+rd.$$ Thus $$\sum_{r=0}^{10}(15+rd)^2=990.$$ Expand: $$\sum_{r=0}^{10}(225+30rd+r^2d^2)=990.$$ So $$11\cdot 225+30d\sum_{r=0}^{10}r+d^2\sum_{r=0}^{10}r^2=990.$$ Using $$\sum_{r=0}^{10}r=55, \qquad \sum_{r=0}^{10}r^2=385,$$ we get $$2475+30d(55)+385d^2=990.$$ That is, $$2475+1650d+385d^2=990,$$ $$385d^2+1650d+1485=0.$$ Divide by $55$: $$7d^2+30d+27=0.$$ Factor: $$(7d+9)(d+3)=0.$$ So $$d=-\frac97 \quad \text{or} \quad d=-3.$$ From the condition $d< -\frac32$, only $$d=-3$$ is valid. --- 3. **Find the mean of the 11 terms** Since the sequence is an AP with first term $15$ and common difference $-3$, $$a_{11}=15+10(-3)=-15.$$ The average of an AP is $$\frac{a_1+a_{11}}{2}= rac{15+(-15)}{2}=0.$$ Therefore, $$\frac{a_1+a_2+\cdots+a_{11}}{11}=0.$$ --- 4. **Compare with stored answer** Derived answer: $0$. Stored correct answer: $0$. They agree.

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