The roots of the quadratic equation 3 x 2 -p x+q=0 are 10 th and 11 th terms of an arithmetic progression with common difference 3/2 . If the sum of the first 11 terms of this arithmetic progression is 88 , then q-2p is equal to \ \ \ \ \

Mathematics · Sequence and Series

JEE Main 2025 — 23 January, Evening Shift — Question 25

The roots of the quadratic equation $3 x^{2}-p x+q=0$ are $10^{\text {th }}$ and $11^{\text {th }}$ 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 $q-2p$ is equal to $\_\_\_\_\_$

Answer: 474

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{S}_{11}=\frac{11}{2}(2 \mathrm{a}+10 \mathrm{~d})=88$

$a+5 d=8$ , $\mathrm{a}=8-5 \times \frac{3}{2}=\frac{1}{2}$

Roots are

$\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=\frac{1}{2}+9 \times \frac{3}{2}=14$

$\mathrm{T}_{11}=\mathrm{a}+10 \mathrm{~d}=\frac{1}{2}+10 \times \frac{3}{2}=\frac{31}{2}$

$\frac{\mathrm{p}}{3}=\mathrm{T}_{10}+\mathrm{T}_{11}=14+\frac{31}{2}=\frac{59}{2}$ $\mathrm{p}=\frac{177}{2}$

$\frac{\mathrm{q}}{3}=\mathrm{T}_{10} \times \mathrm{T}_{11}=7 \times 31=217$

$\mathrm{q}=651$

$q-2 p$

$=651-177$

$=474$

Answer key and solution verified before publishing.

Practise Sequence and Series

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression