NIMCET 2026 Mathematics Question 24 with Solution

Question 24

Mathematics Sequence And Series Hard

If \({{a}}_1,{{a}}_2,\ldots,{{a}}_n\) are any real numbers and \(n\) is any positive integer, then

(A) \(n\sum ^n_{i=1}{{{a}}_i}^2{\lt}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2\)

(B) \(n\sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2\)

(C) \(\sum ^n_{i=1}{{{a}}_i}^2{\geq}{\Bigg{(}\sum ^n_{i=1}{{{a}}_i}\Bigg{)}}^2\)

(D) None of the above

View Dynamic Solution & Explanation ▼

Correct Solution: Option B

Step-by-step Solution:

This question can be solved using a set of numbers like \( 1, 2, 3 \) or \( 1, -2, 3 \). We have: \[ n \sum a_i^2 = 3 \left(1^2 + 2^2 + 3^2 \right) = 42, \] and \[ \left( \sum a_i \right)^2 = (1 + 2 + 3)^2 = 36. \] We see that \[ n \sum a_i^2 \geq \left( \sum a_i \right)^2. \] To prove it algebraically, let us take three numbers \( a, b, c \). We know that: \[ (a - b)^2 + (b - c)^2 + (c - a)^2 \geq 0. \] This implies: \[ a^2 + b^2 + c^2 \geq ab + bc + ca. \] Or: \[ 2 \left( a^2 + b^2 + c^2 \right) \geq 2 (ab + bc + ca). \] Adding \( a^2 + b^2 + c^2 \) to both sides, we have: \[ 3 \left( a^2 + b^2 + c^2 \right) \geq a^2 + b^2 + c^2 + 2 (ab + bc + ca). \] Hence, \[ 3 \left( a^2 + b^2 + c^2 \right) \geq (a + b + c)^2. \]