Mathematics · Sequence and Series

JEE Main 2024 — 30 January, Shift 1 — Question 2

Let $\mathrm{S}_{\mathrm{a}}$ denote the sum of first n terms an arithmetic progression. If $S_{20}=790$ and $S_{10}=145$ , then $S_{15}-$ $\mathrm{S}_{5}$ is :

Option A:
395
Correct
Option B:
390
Option C:
405
Option D:
410

Answer: A

Step-by-step solution

$\mathrm{S}_{20}=\frac{20}{2}[2 \mathrm{a}+19 \mathrm{~d}]=790$

$2 a+19 d=79$

$\mathrm{S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145$

$2 \mathrm{a}+9 \mathrm{~d}=29$

From (1) and (2) $a=-8, d=5$

$\mathrm{S}_{15}-\mathrm{S}_{5}=\frac{15}{2}[2 \mathrm{a}+14 \mathrm{~d}]-\frac{5}{2}[2 \mathrm{a}+4 \mathrm{~d}]$

$=\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20]$

$=405-10$ $=395$

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 2024
Paper: 30 January, Shift 1
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression

Let alpha=1^2+4^2+8^2+13^2+19^2+26^2+ldots ldots upto 10 terms and beta=sum n=1^10 n^4 . If 4 alpha-beta=55 k+40 , then k is equal to

Answer: A

Step-by-step solution

Practise Sequence and Series

More Sequence and Series questions from this paper