Selection of Terms in Geometric Progression

Sometimes we need to assume certain number of terms in Geometric Progression. The following ways are generally used for the selection of terms in Geometric Progression.

(i) If the product of three numbers in Geometric Progression be given, assume the numbers as \(\frac{a}{r}\), a and ar. Here common ratio is r.

(ii) If the product of four numbers in Geometric Progression be given, assume the numbers as \(\frac{a}{r^{3}}\)\(\frac{a}{r}\), ar and ar\(^{3}\). Here common ratio is r\(^{2}\).

(iii) If the product of five numbers in Geometric Progression be given, assume the numbers as \(\frac{a}{r^{2}}\)\(\frac{a}{r}\), a, ar and ar\(^{2}\). Here common ratio is r.

(iv) If the product of the numbers is not given, then the numbers are taken as a, ar, ar\(^{2}\), ar\(^{3}\), ar\(^{4}\), ar\(^{5}\), .....................

 

Solved examples to observe how to use the selection of terms in Geometric Progression:

1. Sum and product of three numbers of a geometric progression are 38 and 1728 respectively. Find the numbers.

Solution:

Let the numbers be \(\frac{a}{r}\), a and ar. Then,

Product = 1728

⇒ \(\frac{a}{r}\) ∙  ar = 1728

⇒ a = 12

Sum = 38

⇒ \(\frac{a}{r}\) + a + ar = 38

⇒ a(\(\frac{1}{r}\) + 1 + r) = 38

⇒ 12(1 + r + \(\frac{r^{2}}{r}\)) = 38

⇒ 6 + 6r + 6r\(^{2}\) = 19r

⇒ 6r\(^{2}\) - 13r + 6 = 0

⇒ (3r - 2)(2r - 3) = 0

⇒ (3r - 2) = 0 or, (2r - 3) = 0

⇒ 3r = 2 or, 2r = 3

⇒ r = \(\frac{2}{3}\) or, r = \(\frac{3}{2}\)

Hence, putting the values of a and r, the required numbers are 8, 12, 18 (Taking r = \(\frac{2}{3}\))

or, 18, 12, 8 (Taking r = \(\frac{3}{2}\))


2. Find three numbers in Geometric Progression whose sum is 35 and product is 1000.

Solution:

Let the required numbers in Geometric Progression be \(\frac{a}{r}\), a and ar.

By the conditions of the problem, we have,

\(\frac{a}{r}\)  a ∙ ar = 1000

a\(^{3}\) = 1000

a = 10 (Since, a is real)

and \(\frac{a}{r}\) + a + ar = 35

a + ar + \(\frac{ar^{2}}{r}\) = 35

10(1 + r + r\(^{2}\)) = 35r (Since a = 10)

2 (1 + r + r\(^{2}\)) = 7r

2 + 2r + 2r\(^{2}\) - 7r = 0

2r\(^{2}\) - 5r + 2 = 0

2r\(^{2}\) - 4r - r + 2 = 0

2r(r - 2) -1(r - 2) = 0

(r - 2)(2r - 1) = 0

Therefore, r = 2 or, ½

Hence, putting the values of a and r, the required numbers are \(\frac{10}{2}\), 10, 10  2 i.e., 5, 10, 20 (Taking r = 2)

Or, 10  2, 10, 10  ½ i.e., 20, 10, 5 (taking r = ½).

 Geometric Progression




11 and 12 Grade Math 

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