Question

if the 5th term of a GP is 2 then the product of its first 9 terms is?

Answer 1

@satellite73 @mukushla

Answer 2

@amistre64

Answer 3

your missing an element, or are you spose to keep it generic?

Answer 4

My question is correct the answer should be 512

Answer 5

(a1) * (a1*r) * (a1*r*r ) * (a1*r*r*r) * (a1*r*r*r*r) ...
=2

Answer 6

I'm thinking if it works...
5th term of a GP is 2 => \(ar^4 = 2\) => \(a= \frac{2}{r^4}\)

General term of a GP = \(ar^{n-1}\)
product of its first 9 terms
= \((ar^{1-1})\times (ar^{2-1}) \times ...\times (ar^{9-1})\)
= \(a^9r^{0+1+2+...+8}\)
= \((\frac{2}{r^4})^9 r^{36}\)
= \(2^9\)

Answer 7

callistos is fancier :)

Answer 8

thxxx

Answer 9

@Yahoo! Welcome.... Do you understand my workings???

Answer 10

yup!!

Answer 11

Glad to hear that :)

Answer 12

@Callisto doubt

Answer 13

Yes?

Answer 14

hw r^34?

Answer 15

It's r^(36)
Since it's r^(0+1+2+..+8)
0+1+2+...+8 = (1+8)x8 / 2 = 36
So, r^(0+1+2+..+8) = r^36

Answer 16

got it thxx

Answer 17

Welcome!!~