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 $b^2 = ac$.Applying this condition: $(x + 14)^2 = (x + 2)(x + 62)$.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 form a $G.P.$ with a common ratio of $4$.

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

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + .....$ is

If $a_1, a_2, a_3, . . . , a_n, . . .$ are in $A.P.$ such that $a_4 - a_7 + a_{10} = m$,then the sum of the first $13$ terms of this $A.P.$ is .............. $m$.

The value of $\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to

The natural numbers are written as follows. The sum of the numbers in the $n^{th}$ row is:
$1$
$2$ $3$
$4$ $5$ $6$
$7$ $8$ $9$ $10$
$. . .$

If $S$ is the sum to infinity of a $G.P.$,whose first term is $a$,then the sum of the first $n$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module