Question

the sum of first three terms of an increasing G.P is 7 and the sum of their squares is 21. Determine the first five terms of the G.P

Answer 1

HELLO DEAR,

Let the first term of GP be a and the common ratio be r .

then, sum of first three terms is 

a + ar + ar² = 7

a(1 + r + r²) = 7----------( 1 )

and,

sum of the square  = 21

therefore, a² + (ar)² + (ar²)² = 21

a²(1 + r² + r⁴) = 21-----------( 2 )

we know:-

(1 + a + a²)(1 - a + a²) = (1 + a²)² - a²

= 1 + a⁴ + 2a² - a²

= (1 + a⁴ + a²)

use it here's,

from----------( 2 )

a[a(1 + r + r²)(1 - r + r²)] = 21

a[7 * (1 - r + r2)] = 21  from----------( 1 )

a(1 - r + r²) = 3-----------( 3 )

dividing--------( 1 ) &--------( 3 )

$\bold{{a(1 + r + r^2) \over a(1 - r + r^2)} = {7\over3}}$

$\bold{\Rightarrow 3(1 + r + r^2) = 7(1 - r + r^2)}$

3 + 3r + 3r² = 7 - 7r + 7r²

4r² - 10r + 4 = 0

4r² - 8r - 2r + 4 = 0

4r(r - 2) - 2(r - 2) = 0

(4r - 2)(r - 2) = 0

r = 1/2 , 2

But r ≠ 1/2 because g.p. is increasing

So,
put r = 2, in -----( 1 )

a = 7/(1 + 2 + 4)

a = 7/7

a = 1,

so, g.p; 1 , 2 , 4 , 8 , 16..........

2......

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Let first three terms of an increasing GP : a/r , a , ar

sum of three terms = 7
=> a/r + a + ar = 7
=> a[1/r + 1 + r] = 7
=> a/r[1 + r + r²] = 7.............(1)

sum of their squares = 21
=> (a/r)² + a² + (ar)² = 21
=> a²/r² + a² + a²r² = 21
=> a² [1/r² + 1 + r²] = 21
=> a²/r² [1 + r² + r⁴ ] = 21

from equation (1),
=> {7/(1 + r + r²) }² × [1 + r² + r⁴] = 21
=> 49(1 + r² + r⁴) = 21(1 + r + r²)²
=> 49(1 + r² - r)(1 + r² + r) = 21(1 + r + r²)²
=> 49(1 + r² - r) = 21(1 + r + r²)
=> 49 + 49r² - 49r = 21 + 21r + 21r²
=> 28 + 28r² - 70r = 0
=> 2r² - 5r + 2 = 0
=> 2r² - 4r - r + 2 = 0
=> 2r(r - 2) -1(r - 2) = 0
=> (2r - 1)(r - 2) = 0
r = 2 , 1/2 but r ≠ 1/2 because GP is in increasing.

now put r = 2 in equation (1),
a/2[1 + 2 + 2²] = 7
a = 2

so, first five terms in GP : a/r, a , ar, ar² , ar³
1 , 2, 4 , 8, 16