Let the first three terms 2, p and q,with q n | vedclass

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

Question

(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$ and the $G.P.$ be $2, p, q, \dots$Given that $2, p, q$ are the $7^{	ext{th}}, 8^{	ext{th}}, 13^{	ext{th

Vedclass · Accepted Answer

$163$

Vedclass · Answer

(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$ and the $G.P.$ be $2, p, q, \dots$Given that $2, p, q$ are the $7^{	ext{th}}, 8^{	ext{th}}, 13^{	ext{th}}$ terms of the $A.P.$:$2 = a + 6d \quad \dots(i)$$p = a + 7d \quad \dots(ii)$$q = a + 12d \quad \dots(iii)$Subtracting $(i)$ from $(ii)$,we get $p - 2 = d$.Subtracting $(ii)$ from $(iii)$,we get $q - p = 5d$.Since $q - p = 5(p - 2)$,we have $q = 6p - 10$.Since $2, p, q$ are in $G.P.$,$p^2 = 2q$.Substituting $q = 6p - 10$,we get $p^2 = 2(6p - 10) \implies p^2 - 12p + 20 = 0$.Solving for $p$,$(p - 10)(p - 2) = 0$,so $p = 10$ or $p = 2$.If $p = 2$,then $q = 2$,which contradicts $q 
eq 2$. Thus,$p = 10$.Then $d = p - 2 = 8$ and $a = 2 - 6(8) = -46$.The $G.P.$ is $2, 10, 50, 250, 1250, \dots$The $5^{	ext{th}}$ term of the $G.P.$ is $2 	imes 5^4 = 1250$.Let this be the $n^{	ext{th}}$ term of the $A.P.$: $1250 = a + (n - 1)d$.$1250 = -46 + (n - 1)8 \implies 1296 = (n - 1)8 \implies n - 1 = 162 \implies n = 163$.

Let the first three terms $2, p$ and $q$,with $q \neq 2$,of a $G.P.$ be respectively the $7^{\text{th}}$,$8^{\text{th}}$ and $13^{\text{th}}$ terms of an $A.P.$ If the $5^{\text{th}}$ term of the $G.P.$ is the $n^{\text{th}}$ term of the $A.P.$,then $n$ is equal to

Similar Questions

If the arithmetic mean and geometric mean of two positive numbers are $A$ and $G$ respectively,then the numbers are ..........

The Geometric Mean $(G.M.)$ and Harmonic Mean $(H.M.)$ of two numbers are $10$ and $8$ respectively. The numbers are:

Three non-zero real numbers form an $A.P.$ and the squares of these numbers taken in the same order form a $G.P.$ Then the number of all possible common ratios of the $G.P.$ is

If $a, b, c$ are any three positive numbers,then what is the minimum value of $(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$?

If the ratio of the harmonic mean to the geometric mean of two positive numbers $a$ and $b$ $(a > b)$ is $4 : 5$,then $a : b = \dots$ (in $: 1$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module