JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Sequences and Series

The roots of the quadratic equation

3 x^{2} - p x + q = 0

are

1 0^{th}

and

1 1^{th}

terms of an arithmetic progression with common difference

\frac{3}{2}

. If the sum of the first 11 terms of this arithmetic progression is 88, then

q - 2 p

is equal to ______.

Answer: 474

Solution

Let the AP have first term

a

and common difference

d = \frac{3}{2}

. Since

S_{11} = 88

S_{11} = \frac{11}{2} (2 a + 10 d) = 88 \Rightarrow 2 a + 10 d = 16 \Rightarrow a + 5 d = 8

a + 5 (\frac{3}{2}) = 8 \Rightarrow a = \frac{1}{2}

The 10th and 11th terms:

T_{10} = a + 9 d = \frac{1}{2} + \frac{27}{2} = 14, T_{11} = a + 10 d = \frac{1}{2} + 15 = \frac{31}{2}

These are the roots of

3 x^{2} - p x + q = 0

, so

\frac{p}{3} = T_{10} + T_{11} = 14 + \frac{31}{2} = \frac{59}{2} \Rightarrow p = \frac{177}{2}

\frac{q}{3} = T_{10} \cdot T_{11} = 14 \times \frac{31}{2} = 217 \Rightarrow q = 651

q - 2 p = 651 - 177 = 474

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Integer answer