Question

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}}
a(1−r+r
2
)
a(1+r+r
2
)
​
=
3
7
​


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

Answer 1

Answer:

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 )

Answer 2

Answer:

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