Here we will learn how to solve different types of problems on Geometric Progression.
1. Find the common ratio of the Geometric Progression whose, sum of the third and fifth terms is 90 and its first term is 1.
The first term of the given Geometric Progression a = 1.
Let ‘r’ be the common ratio of the Geometric Progression.
According to the problem,
t_3 + t_5 = 90
ar^2 + ar^4 = 90
r^2 + r^4 = 90
r^4 + r^2 – 90 = 0
r^2 + 10r^2 - 9r^2 - 90 = 0
(r^2 + 10)(r^2 - 9) =0
r^2 - 9 = 0
r^2 = 9
r = ±3
Therefore, the common ratio of the Geometric Progression is -3 or 3.
2. Find a Geometric Progress for which the sum of first two terms
is -4 and the fifth term is 4 times the third term.
Let ‘a’ be the first term and ‘r’ be the common ratio of the given Geometric Progression.
Then, according to the problem the sum of first two terms is -4
t_1 + t_2 = -4
a + ar = -4 .................. (i)
and the fifth term is 4 times the third term.
t_5 = 4t_3
ar^4 = 4ar^2
r^2 = 4
r = ±2
Putting r = 2 and -2 respectively in (i), we get a = -4/3 and a = 4.
Thus, the required Geometric Progression is -4/3, -8/3, -16/3, ............ or 4, -8, 16, -32, ........................
3. Prove that in a Geometric Progression of finite number of terms the product of any two terms equidistant from the beginning and the end is constant and is equal to the product of the first and last and last terms.
Let ‘a’ be the first term, ‘b’ the last term and ‘r’ the common ratio of a finite Geometric Progression.
Then the nth term from the beginning = a* r^(n - 1)
And the nth term from the end = b/r^(n -1)
Therefore, the product of two equidistant terms from the beginning and end (i.e., the terms at the nth positions) = a * r^(n - 1) * b/r^(n -1) = a * b = constant = first term X last term. Proved.
● Geometric Progression
11 and 12 Grade Math
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