Question

The first term of a geometric progression is 1. The sum of the third and fifth term is 90. Find the common ratio of the geometric progression.

Answer 1

$T_1 = a = 1 \\ \\ \\ T_3 + T_5 = 90 \\ \\ = ar^2 + ar^4 = 90 \\ \\ = (1 \times r^2) + (1 \times r^4) = 90 \\ \\ = r^2 + r^4 = 90 \\ \\ = r^2(1 + r^2) = 90 = 9 \times 10 = 9(1 + 9) = 3^2(1 + 3^2) \\ \\ \\ \therefore\ r = \bold{3}$

$\bold{3}\ is\ the\ common\ ratio.$

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Answer 2

Let the terms be a,ar^2,ar^3,ar^4,ar^5.

Now,

Given that first term of the GP be 1.

a = 1.

Now,

We know that nth term of GP tn = a * r^n - 1

= > t3 = a * r^3 - 1 = ar^2

= > t5 = a * r^5 - 1 = ar^4.

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Given that sum of third term and fifth term is 90.

= > ar^2 + ar^4 = 90

= > a(r^2 + r^4) = 90

= > 1(r^2 + r^4) = 90

= > r^2 + r^4 = 90

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

= > r^4 + 10r^2 - 9r^2 - 90 = 0

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

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

= > r^2 - 9 = 0, r^2 = -10[neglect -ve values]

= > r^2 = 9

= > r = +3,-3.

Therefore, the common ratio of the geometric progression is +3,-3.

Hope this helps!