If the third term of a G.P. is 4,then the pro | vedclass

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

Question

(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.The terms of the $G.P.$ are $a, ar, ar^2, ar^3, ar^4, \dots$Given that the thi

Vedclass · Accepted Answer

$4^5$

Vedclass · Answer

(C) Let the first term of the $G.P.$ be $a$ and the common ratio be $r$.The terms of the $G.P.$ are $a, ar, ar^2, ar^3, ar^4, \dots$Given that the third term is $4$,so $ar^2 = 4$.The product of the first $5$ terms is $P = a 	imes (ar) 	imes (ar^2) 	imes (ar^3) 	imes (ar^4)$.$P = a^5 	imes r^{(1+2+3+4)} = a^5 	imes r^{10}$.$P = (ar^2)^5$.Substituting the value $ar^2 = 4$,we get $P = 4^5$.

If the third term of a $G.P.$ is $4$,then the product of its first $5$ terms is:

Similar Questions

The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is

If ${S_n} = nP + \frac{1}{2}n(n - 1)Q$,where ${S_n}$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$ is

The sum of all two-digit positive numbers which, when divided by $7$, yield $2$ or $5$ as a remainder is:

Find the sum of the first hundred even natural numbers divisible by $5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module