The number which should be added to the numbe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the number to be added be $x$.Then,the numbers $x + 2, x + 14, x + 62$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the number to be added be $x$.Then,the numbers $x + 2, x + 14, x + 62$ are in $G.P.$For three numbers $a, b, c$ to be in $G.P.$,the condition is $b^2 = ac$.Therefore,$(x + 14)^2 = (x + 2)(x + 62)$.Expanding both sides: $x^2 + 28x + 196 = x^2 + 64x + 124$.Subtracting $x^2$ from both sides: $28x + 196 = 64x + 124$.Rearranging the terms: $196 - 124 = 64x - 28x$.$72 = 36x$.$x = 2$.Thus,adding $2$ to the given numbers results in $4, 16, 64$,which are in $G.P.$ as $16^2 = 4 	imes 64$ $(256 = 256)$.

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$ is

Similar Questions

If the fifth term of a $G.P.$ is $2$,then the product of its first $9$ terms is:

The geometric mean of the sequence $1, 2, 2^2, ..., 2^n$ is:

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

What is the geometric mean of the observations $2, 4, 8, 16, 32, 64$?

If $25, x - 6$,and $x - 12$ are consecutive terms of a geometric progression,then $x = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module