Let three real numbers a, b, c be in arithmetic progression and a +1, b , c +3 be in geometric progression. If a 10 and the arithmetic mean of a , b and c is 8 , then the cube of the geometric mean of a, b and c is

Mathematics · Sequence and Series

JEE Main 2024 — 4 April, Shift 2 — Question 5

Let three real numbers $a, b, c$ be in arithmetic progression and $\mathrm{a}+1, \mathrm{~b}, \mathrm{c}+3$ be in geometric progression. If $\mathrm{a}>10$ and the arithmetic mean of $\mathrm{a}, \mathrm{b}$ and c is 8 , then the cube of the geometric mean of $a, b$ and $c$ is

Option A:
120
Correct
Option B:
312
Option C:
316
Option D:
128

Answer: A

Step-by-step solution

$2 \mathrm{~b}=\mathrm{a}+\mathrm{c}, \mathrm{b}^{2}=(\mathrm{a}+1)(\mathrm{c}+3)$ ,

$\frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16$

$64=(a+1)(19-a)=19+18 a-a^{2}$

$a^{2}-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10)$

$\mathrm{a}=15, \mathrm{c}=1, \mathrm{~b}=8$

$\left((\mathrm{abc})^{1 / 3}\right)^{3}=\mathrm{abc}=120$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Sequence and Series
Topic: Geometric Progression

The value of frac1 × 2^2+2 × 3^2+ldots+100 ×(101)^21^2 × 2+2^2 × 3+ldots+100^2 × 101 is

Answer: A

Step-by-step solution

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