If for the series a_1, a_2, a_3, ldots,the ex | vedclass

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

Question

(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).Dividing the terms,we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{a

Vedclass · Accepted Answer

$H$.$P$.

Vedclass · Answer

(C) Given that $\frac{a_r - a_{r+1}}{a_r a_{r+1}} = K$ (where $K$ is a constant).Dividing the terms,we get $\frac{a_r}{a_r a_{r+1}} - \frac{a_{r+1}}{a_r a_{r+1}} = K$.This simplifies to $\frac{1}{a_{r+1}} - \frac{1}{a_r} = K$.Since the difference between the reciprocals of consecutive terms is constant,the sequence $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots$ is an $A$.$P$.Therefore,the sequence $a_1, a_2, a_3, \ldots$ is in $H$.$P$.

If for the series $a_1, a_2, a_3, \ldots$,the expression $a_r - a_{r+1}$ bears a constant ratio with $a_r a_{r+1}$,then $a_1, a_2, a_3, \ldots$ are in:

Similar Questions

If $a, b, c$ are in $H.P.$,then the value of $\left( \frac{1}{b} + \frac{1}{c} - \frac{1}{a} \right) \left( \frac{1}{c} + \frac{1}{a} - \frac{1}{b} \right)$ is

If the sixth term of a $H.P.$ is $\frac{1}{61}$ and its tenth term is $\frac{1}{105},$ then the first term of that $H.P.$ is

Which number should be added to the numbers $13, 15, 19$ so that the resulting numbers are the consecutive terms of a $H.P.$?

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

Find the harmonic mean of $a/b$ and $b/a$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module