Question

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

Answer 1

Answer:

216

Step-by-step explanation:

Let the three consecutive numbers in the GP be: $a/r,a,ar$.

The first given is that the sum of those three consecutive number is 26.

Therefore, $\frac{a}{r}+a+ar=26$

$a(\frac{1}{r}+1+r)=26\\\frac{a}{r} (1+r+r^2)=26$

Square both sides:

$\frac{a^2}{r^2} (1 + r + r^2)^2 = 676$ (1)

From the second given, the sum of the squares of the three numbers is 364:

Hence,

$\frac{a^2}{r^2}+a^2+a^2r^2=364\\a^2(\frac{1}{r^2}+1+r^2)=364\\\frac{a^2}{r^2}(1+r^2+r^4)=364\\\frac{a^2}{r^2}(1-r+r^2)=364\:\:\:(2)$

From equations (1) and (2):

$\frac{(1-r+r^2)}{(1+r+r^2)}=\frac{364}{676}\\\frac{(1-r+r^2)}{(1+r+r^2)}=\frac{7}{13}\\13(1-r+r^2)=7(1+r+r^2)\\13 - 13r + 13r^2 = 7 + 7r + 7r^2\\6 - 20r + 6r^2 = 0\\3r^2 - 10r + 3 = 0\\(3r - 1)(r - 3) = 0\\\\r = 3\:\:\:or\:\:\:r=1/3$

Substitute with r=3 in eqn. (1):

$a(1 + 3 + 3^2)/3 = 26\\a*13/3 = 26\\a=6$

Therefore, the product of these numbers: $a/r*a*ar=a^3=6^3=216$

Answer 2

Answer:

216

Step-by-step explanation:

let say three numbers are

a  , ar , ar^2

a + ar + ar^2 = 26    eqA

a^2 + a^r^2 + a^2r^4 = 364   eqB

Squaring EqA

a^2 + 2a^2r + 3a^2r^2 + 2a^2r^3 + a^2r^4 = 676    - eq C

Eq C - Eq B

2a^2r + 2a^2r^2 +2a^2r^3 = 676-314

a^2r + a^2r^2 + a^r^3 = 156

ar * (a + ar + ar^2) = 156

ar * 26 = 156   from Eq A

ar = 6

Products of three numbers

a * ar * ar^2 =  a^3r^3 = (ar)^3

= 6^3 = 216