AP ICET 2019 Shift 1 Paper

Question : 128 of 200

Marks: +1, -0

If the sum of three consecutive numbers in a geometric progression is 26 and the sum of their squares is 364. then the product of those numbers is

Solution:

Let 3 terms of a G.P be

a/r,a,ar

a({1}/{r}+1+r)=26

...(1)
also

a^{2}(\;{1}/{r^{2}}+1+r^{2})=364

...(2)
Squaring on both sides (1)

a^{2}(\;{1}/{r^{2}}+1+r^{2}+2(\;{1}/{r}+1+r))=676

a^{2}(\;{1}/{r^{2}}+1+r^{2})+2 a(a(\;{1}/{r}+1+r))=676

364 + 2a (26) = 676⇒ a =6

Product of the terms is

a^3 = 6^3 = 216

Go to Question:

Prev Question

Next Question