# The first term of a G.P. is 1. The sum of the third and fifth term is 90. Calculate the common ratio of the G.P.

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a1, a2, a3, ..... is a geometric progression.

We know that:

a1= 1

a3 + a5 = 90

But:

a3= a1*r^2 = r^2

a5 = a1 *r^4 = r^4

==> r^2 + r^4 = 90

==> r^4 + r^2 - 90 =0

==> (r^2 + 10) (r^2 - 9) = 0

==> (r^2 + 10) ( r-3)(r+3) = 0

==> r1= 3

==> r2= -3

We'll note the first term of the G.P. as a1 = 1 and the common ratio as r.

From the enunciation, we know that:

a3 + a5 = 90

From the standard formula of the n-th term of the G.P., we have:

an = a1*r^(n-1)

Now, we can calculate a3 and a5:

a3 = a1*r^2 = r^2

a5 = a1*r^4 = r^4

a3 + a5 = 90 => r^2 + r^4 = 90

We'll substitute r^2 = t

t^2 + t - 90 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1+360)]/2

t1 = (-1+19)/2

t1 = 9

t2 = (-1-19)/2

t2 = -10

But r^2 = t

So, r^2 = t1

r^2 = 9

r1 = +3

r2 = -3

r^2 = t2

r^2 = -10 impossible

So, the common ratio could be r = -3 or r = 3.