If the {p^{th}},;{q^{th}} and {r^{th}} term o | vedclass

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

Question

(b) Let $A$ and $R$ be the first term and common ratio of the $G.P.$

Then $a = A{R^{p - 1}},\;b = A{R^{q - 1}}$ and $c = A{R^{r - 1}}$&hellip;..(i)

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(b) Let $A$ and $R$ be the first term and common ratio of the $G.P.$

Then $a = A{R^{p - 1}},\;b = A{R^{q - 1}}$ and $c = A{R^{r - 1}}$&hellip;..(i)

Again if $x$ and $d$ be the first and common difference of the $A.P.$ corresponding to the given $H.P.$

Then $\frac{1}{a} = x + (p - 1)d,\;\frac{1}{b} = x + (q - 1)d$,

$\frac{1}{c} = x + (r - 1)d$&hellip;..(ii)

From (i), $\frac{a}{b} = {R^{p - q}}$

or ${\left( {\frac{a}{b}} ight)^{1/c}} = {({R^{p - q}})^{1/c}} = {R^k}$, where $k = \frac{{p - q}}{c}$

From (ii), $k = (p - q)\left\{ {x + (r - 1)d} ight\}$

$ = (p - q)x - (p - q)(r - 1)d$

$ = (p - q)x - (p - q)d + (rp - rq)d$&hellip;..(iii)

Similarly, ${\left( {\frac{b}{c}} ight)^{1/a}} = {({R^{q - r}})^{1/a}} = {R^n}$, where $n = \frac{{q - r}}{a}$

$ \Rightarrow $$n = (q - r) 	imes \left\{ {x + (p - 1)d} ight\}$

$ \Rightarrow $$n = (q - r)x - (q - r)d + (pq - pr)d$&hellip;..(iv)

and ${\left( {\frac{c}{a}} ight)^{1/b}} = {({R^{r - p}})^{1/b}} = {R^m}$

Where $m = \frac{{r - p}}{b} = (r - p)\left\{ {x + (q - 1)d} ight\}$

$ = (r - p)x - (r - p)d + (rq - qp)d$&hellip;..(v)

Hence ${\left( {\frac{a}{b}} ight)^{1/c}}{\left( {\frac{b}{c}} ight)^{1/a}}{\left( {\frac{c}{a}} ight)^{1/b}} = {R^k}{R^m}{R^n} = {R^{m + n + k}}$

$ = {R^0} = 1${since $k + m + n = 0$}, Adding (iii), (iv), (v)

Taking logarithm of both sides, we get

$\frac{1}{c}({\log _e}a - {\log _e}b) + \frac{1}{a}({\log _e}b - {\log _e}c)$

$+ \frac{1}{b}({\log _e}c - {\log _e}a) = {\log _e}(1)$

$ \Rightarrow $$\left( {\frac{1}{c} - \frac{1}{b}} ight){\log _e}a + \left( {\frac{1}{a} - \frac{1}{c}} ight){\log _e}b + \left( {\frac{1}{b} - \frac{1}{a}} ight){\log _e}c = 0$

$ \Rightarrow $$\left( {\frac{{b - c}}{{bc}}} ight){\log _e}a + \left( {\frac{{c - a}}{{ac}}} ight){\log _e}b + \left( {\frac{{a - b}}{{ab}}} ight){\log _e}c = 0$

$ \Rightarrow $$a(b - c){\log _e}a + b(c - a){\log _e}b + c(a - b){\log _e}c = 0$.

Note : Such type of questions $i.e.$ containing term with cyclic coefficient associated with negative sign reduce to $0$ mostly.

If the ${p^{th}},\;{q^{th}}$ and ${r^{th}}$ term of a $G.P.$ and $H.P.$ are $a,\;b,\;c$, then $a(b - c)\log a + b(c - a)$ $\log b + c(a - b)\log c = $

Similar Questions

If $A.M.$ and $G.M.$ of roots of a quadratic equation are $8$ and $5,$ respectively, then obtain the quadratic equation.

If the first and ${(2n - 1)^{th}}$ terms of an $A.P., G.P.$ and $H.P.$ are equal and their ${n^{th}}$ terms are respectively $a,\;b$ and $c$, then

If $a,\;b,\;c$ are in $H.P.$, then for all $n \in N$ the true statement is

If $m$ is the $A.M$ of two distinct real numbers $ l$ and $n (l,n>1) $ and $G_1, G_2$ and $G_3$ are three geometric means between $l$ and $n$ then $G_1^4 + 2G_2^4 + G_3^4$ equals :

If the arithmetic mean and geometric mean of the $p ^{\text {th }}$ and $q ^{\text {th }}$ terms of the sequence $-16,8,-4,2, \ldots$ satisfy the equation $4 x^{2}-9 x+5=0,$ then $p+q$ is equal to ..... .