Terms a, 1, b are in arithmetic progression a | vedclass

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

Question

(D) Given that $a, 1, b$ are in arithmetic progression $(AP)$.Therefore,$1 = \frac{a+b}{2}$,which implies $a+b = 2$ $......(1)$.Given that $1, a, b$ a

Vedclass · Accepted Answer

$-2, 4$

Vedclass · Answer

(D) Given that $a, 1, b$ are in arithmetic progression $(AP)$.Therefore,$1 = \frac{a+b}{2}$,which implies $a+b = 2$ $......(1)$.Given that $1, a, b$ are in geometric progression $(GP)$.Therefore,$a^2 = 1 	imes b$,which implies $b = a^2$ $......(2)$.Substituting the value of $b$ from equation $(2)$ into equation $(1)$:$a + a^2 = 2$$a^2 + a - 2 = 0$Factoring the quadratic equation:$(a+2)(a-1) = 0$So,$a = -2$ or $a = 1$.If $a = 1$,then $b = (1)^2 = 1$. However,the problem states $a 
eq b$,so we reject this case.If $a = -2$,then $b = (-2)^2 = 4$.Thus,the values are $a = -2$ and $b = 4$.

Terms $a, 1, b$ are in arithmetic progression and terms $1, a, b$ are in geometric progression. Find $a$ and $b$ (given $a \neq b$).

Similar Questions

Write the seventh term of the following $GP$: $6, 12, 24, 48, \ldots$

The sum of a series in $A.P.$ is $525$. Its first term is $3$ and last term is $39$. Find the common difference.

If $x > 1, y > 1, z > 1$ are in $G.P.$,then $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in

The product of three geometric means between $4$ and $\frac{1}{4}$ will be

If $p, q, r$ are in $A.P.$ and are positive,the roots of the quadratic equation $px^2 + qx + r = 0$ are all real for

Vedclass Test Series

Exam Paper Generator

Online Exam Module