the sum of three numbers in a gp is 14 and the sum of their squares is 84. find the largest number
Answers
Let the first term of the gp be a then :
a, ar, ar²
Their sum :
a + ar + ar² = 14.... 1
a( 1 + r + r²) = 14
Their squares :
a² + a²r² + a²r⁴= 84..... 2
a²(1 + r² + r⁴) = 84
Now squaring 1 we have :
a² + a²r² + a²r⁴ + 2( a²r + a²r³ + a²r²) = 196..... 3
Now subtracting 2 from 3 we have
2(a²r + a²r² + a²r³) = 196 - 84 = 112
2r (a² + a²r + a²r²) = 112
r (a² + a²r + a²r²) = 112/2
r (a² + a²r + a²r²) = 56
ar(a + ar + ar²) = 56
The brackets components equals to equation 1.
Hence :
ar(14) = 56
ar = 56/14
ar = 4
a = 4/r
From the first equation :
a(1 + r + r²) = 14
4/r(1 + r + r²) = 14
1/r(1 + r + r²) = 14/4
1/r(1 + r + r²) = 3.5
1 + r + r² = 3.5r
r² - 2.5r + 1 = 0
Multiply through by 4
4r² - 10r + 4 = 0
4r² - 2r - 8r + 4 = 0
2r(2r - 1) - 4(2r - 1) = 0
(2r - 4)(2r - 1) = 0
2r = 4
r = 2
2r = 1
r = 0.5
We can either use 0.5 or 2
If we take 2.
a = 4/2 = 2
ar = 2 × 2 = 4
ar² = 8
If we take 0.5 we will have the same values but now ordered differently.
So the largest number is 8.
Answer:
I hope this is the correct answer
