If a, b, c are in AP as well as in GP, then (a) a=b ≠c (b) a ≠b=c (c) a ≠b ≠c (d) a=b=c

Name: Problem 65, Chapter 5: IIT-JEE Super Course in Mathematics
Uploaded: 2021-07-03T18:13:02-08:00
Duration: 1 min 6 s
Channel: Narayan Hari
Description: If a, b, c are in AP as well as in GP, then (a) a=b ≠c (b) a ≠b=c (c) a ≠b ≠c (d) a=b=c

step 1 In this problem I can write the value of 2b is equal to a plus c the value of b squared is equal

If a, b, c are in AP as well as in GP, then (a) a=b ≠c (b) a...

If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in AP as well as in GP, then
(a) $\mathrm{a}=\mathrm{b} \neq \mathrm{c}$
(b) $\mathrm{a} \neq \mathrm{b}=\mathrm{c}$
(c) $\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$
(d) $\mathrm{a}=\mathrm{b}=\mathrm{c}$

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence in which the difference between consecutive terms is constant. This means any three terms in an AP satisfy the property that the middle term is the average of the first and third terms.

Geometric Progression

A geometric progression (GP) is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio. For any three terms in a GP, the square of the middle term equals the product of the first and third terms.

Simultaneous Conditions for AP and GP

When three numbers are required to be in both an arithmetic and a geometric progression, the conditions from both sequences must be satisfied simultaneously. The AP condition forces the middle term to be the average of the first and third terms, while the GP condition requires that the square of the middle term equals the product of the first and third, which ultimately leads to the conclusion that all three terms must be equal.

Transcript

Please give Ace some feedback