Let a, b and c be in G.P. with common ratio r | vedclass

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

Question

(D) Given $a, b, c$ are in $G.P.$ with common ratio $r,$ so $b = ar$ and $c = ar^2.$Since $3a, 7b, 15c$ are in $A.P.,$ the middle term property gives

Vedclass · Accepted Answer

$a$

Vedclass · Answer

(D) Given $a, b, c$ are in $G.P.$ with common ratio $r,$ so $b = ar$ and $c = ar^2.$Since $3a, 7b, 15c$ are in $A.P.,$ the middle term property gives $2(7b) = 3a + 15c.$Substituting $b$ and $c,$ we get $14(ar) = 3a + 15(ar^2).$Since $a 
e 0,$ we divide by $a$ to get $15r^2 - 14r + 3 = 0.$Factoring the quadratic equation: $(3r - 1)(5r - 1) = 0,$ which gives $r = \frac{1}{3}$ or $r = \frac{1}{5}.$Given $0 The common difference $d = 7b - 3a = 7ar - 3a = a(7r - 3).$For $r = \frac{1}{3},$ $d = a(7/3 - 3) = -\frac{2}{3}a.$The $4^{th}$ term is $15c + d = 15ar^2 - \frac{2}{3}a = 15a(1/9) - \frac{2}{3}a = \frac{5}{3}a - \frac{2}{3}a = a.$

Let $a, b$ and $c$ be in $G.P.$ with common ratio $r,$ where $a \ne 0$ and $0 < r \le \frac{1}{2}.$ If $3a, 7b$ and $15c$ are the first three terms of an $A.P.,$ then the $4^{th}$ term of this $A.P.$ is

Similar Questions

The first two terms of a geometric progression add up to $12.$ The sum of the third and the fourth terms is $48.$ If the terms of the geometric progression are alternately positive and negative,then the first term is

Three numbers are selected from the set ${3^1, 3^2, 3^3, \dots, 3^{20}}$. Find the number of ways these selected numbers can form an increasing Geometric Progression $(G.P.)$.

The sum of numbers from $250$ to $1000$ which are divisible by $3$ is

If $b$ is the first term of an infinite $G.P.$ whose sum is $5$,then $b$ lies in the interval

Sum the series to infinity $\frac{3}{4} - \frac{5}{4^2} + \frac{3}{4^3} - \frac{5}{4^4} + \frac{3}{4^5} - \frac{5}{4^6} + \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module