The sum of three consecutive terms in a geome | vedclass

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

Question

(B) Let the three consecutive terms of the geometric progression be $a/r, a, ar$.Given that their sum is $14$,so $a/r + a + ar = 14$,which implies $a(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the three consecutive terms of the geometric progression be $a/r, a, ar$.Given that their sum is $14$,so $a/r + a + ar = 14$,which implies $a(1/r + 1 + r) = 14$ ..... $(i)$According to the problem,if $1$ is added to the first and second terms and $1$ is subtracted from the third,the new terms are $a/r + 1, a + 1, ar - 1$.Since these are in arithmetic progression,$2(a + 1) = (a/r + 1) + (ar - 1)$.$2a + 2 = a/r + ar$.$2a + 2 = a(1/r + r)$.From $(i)$,$a(1/r + r) = 14 - a$. Substituting this into the equation:$2a + 2 = 14 - a$.$3a = 12$,so $a = 4$.Substituting $a = 4$ into $(i)$: $4(1/r + 1 + r) = 14$.$1/r + 1 + r = 3.5$.$1/r + r = 2.5$.$r^2 - 2.5r + 1 = 0$.$2r^2 - 5r + 2 = 0$.$(2r - 1)(r - 2) = 0$.So,$r = 2$ or $r = 1/2$.If $r = 2$,the terms are $4/2, 4, 4(2)$,which are $2, 4, 8$.If $r = 1/2$,the terms are $4/(1/2), 4, 4(1/2)$,which are $8, 4, 2$.In both cases,the terms are $2, 4, 8$. The lowest term is $2$.

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

Similar Questions

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,then ${a_1}{a_2} + {a_2}{a_3} + \dots + {a_{n-1}}{a_n}$ is equal to:

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

The sum of an infinite geometric series with positive terms is $3$ and the sum of the cubes of its terms is $\frac{27}{19}$. Then the common ratio of this series is

If the sum of $n$ terms of an $A.P.$ is $2n^2 + 5n$,then the $n^{th}$ term will be

If $x, y, z$ are three positive numbers in $G.P.$,then $\frac{1 + \ln x}{2}, \frac{1 + \ln y}{4}, \frac{1 + \ln z}{8}$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module