In a H.P.,the p^{th} term is q and the q^{th} | vedclass

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

Question

(B) Let the first term of the corresponding $A.P.$ be $A$ and the common difference be $D$. Since the terms of $H.P.$ are the reciprocals of the terms

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let the first term of the corresponding $A.P.$ be $A$ and the common difference be $D$. Since the terms of $H.P.$ are the reciprocals of the terms of $A.P.$,we have:$T_p = \frac{1}{q} \implies A + (p - 1)D = \frac{1}{q} \quad (i)$$T_q = \frac{1}{p} \implies A + (q - 1)D = \frac{1}{p} \quad (ii)$Subtracting $(ii)$ from $(i)$:$(p - q)D = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq}$$D = \frac{1}{pq}$Substituting $D$ in $(i)$:$A + (p - 1)\frac{1}{pq} = \frac{1}{q}$$A = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq}$Now,the $(pq)^{th}$ term of the $A.P.$ is:$T_{pq} = A + (pq - 1)D = \frac{1}{pq} + (pq - 1)\frac{1}{pq} = \frac{1 + pq - 1}{pq} = 1$Since the $(pq)^{th}$ term of the $A.P.$ is $1$,the $(pq)^{th}$ term of the $H.P.$ is the reciprocal of $1$,which is $1$.

In a $H.P.$,the $p^{th}$ term is $q$ and the $q^{th}$ term is $p$. Then the $(pq)^{th}$ term is

Similar Questions

If $\ln(a+c), \ln(c-a), \ln(a-2b+c)$ are in $A.P.$,then

If $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in Arithmetic Progression $(AP)$,then $a^2, b^2, c^2$ are in which progression?

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

If the roots of the equation $x^3 - ax^2 + bx - c = 0$ are in harmonic progression $(HP)$,then the harmonic mean of the roots is

The terms $\log _3 2, \log _6 2, \log _{12} 2$ are in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module