Sequences & Series
Sequences And Series
nta_abhyas_2025
Grade 11
Question:
Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1 = 15, 27 - 2a_2 > 0$ and $a_i = 2a_{i-1} - a_{i-2}$ for $k = 3, 4, \ldots, 11$. If $\frac{a_1 + a_2 + \ldots + a_{11}}{11} = 90$, then the value of $\frac{a_2 + a_4 + \ldots + a_{11}}{11}$ is equal to
Step-by-Step Solution
Key Concept: In an AP, the sum of terms depends on the first term and the sum of deviations, which can be expressed using the common difference.
Given $a_1 = 2a_1 - a_{-2}$, we have $a_{-2} = a_1$. Since $a_1, a_2, \ldots, a_{11}$ are in AP with common difference $d$, we form the equation $\frac{a_1^2 + a_2^2 + \cdots + a_{11}^2}{11} = 90$. Expanding using AP properties: $\frac{11a_1^2 + 2a_1(0 + d + 2d + \cdots + 10d) + (0^2 + 1^2 + \cdots + 10^2)d^2}{11} = 90$. This simplifies to $225 + 35d^2 + 150d = 90$, giving $35d^2 + 150d + 135 = 0$ or $7d^2 + 30d + 27 = 0$. Solving: $d = -3$ or $d = -\frac{9}{7}$. Given $a_2 < \frac{7}{2}$, we have $d = -3$ and $a_1 = 15$. Therefore $a_1 + a_2 + \cdots + a_{11} = 11 \times 15 + (0-3) = 0$.
Correct Answer: 0