# Selection of terms in geometric progression

selection of terms in geometric progression : Sometimes it is required to select the finite number of terms in G.P. It is always easy if we select the terms in the following manner.

 No. of terms Terms common ratio 3 $\frac{a}{r}$, a,ar r 4 $\frac{a}{r^{3}}, \frac{a}{r},ar, ar^{3}$ $r^{2}$ 5 $\frac{a}{r^{2}}, \frac{a}{r},a,ar, ar^{2}$ r

If the product of the numbers is not given, then the numbers are taken as a, ar,a$r^{2}, ar^{3}$,...

## Examples on selection of terms in geometric progression

1) Find the four numbers in G.P whose sum is 85 and product is 4096.
Solution : Let the four numbers in G.P. be $\frac{a}{r^{3}}, \frac{a}{r}, ar, ar^{3}$

Product = 4096
$\frac{a}{r^{3}} \times \frac{a}{r} \times ar\times ar^{3}$ = 4096

∴ $a^{4}$ =4096
$a^{4} = 8^{4}$
∴ a = 8
$\frac{a}{r^{3}} + \frac{a}{r} + ar + ar^{3}$ = 8

a$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85

8$\left (\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3} \right )$ = 85

8$\left ( r^{3} + \frac{1}{r^{3}} \right ) + 8\left ( r + \frac{1}{r} \right )$ = 85

8$\left [ \left ( r + \frac{1}{r} \right )^{3} -3\left(r +\frac{1}{r} \right) \right ]+ 8\left(r + \frac{1}{r} \right)$ = 85
8$\left ( r + \frac{1}{r} \right )^3 - 16\left ( r + \frac{1}{r} \right )$ - 85=0

8$x^{3}$ - 16x - 85 = 0 ----------Let x = r + $\frac{1}{r}$
(2x -5)(4$x^{2}$ + 10 x + 17) = 0
⇒ 2x - 5 = 0 ----- since(4$x^{2}$ + 10 x + 17) has imaginary roots
∴ x = $\frac{5}{2}$
⇒ r + $\frac{1}{r} = \frac{5}{2}$ ⇒ 2$r^{2}$ - 5r + 2 = 0

So r = 2 or r = $\frac{1}{2}$
Put a = 8 and r = 2 or $\frac{1}{2}$ ,we get the four numbers are either 1,4,16,64 or 64,16,4,1

2) Find the three numbers in G.P whose sum is 52 and the sum of whose products in pairs is 624.
Solution : Let the required numbers be a, ar,a$r^{2}$
Sum = 52
⇒ a + ar + a$r^{2}$ = 52
⇒ a( 1 + r + $r^{2}$)= 52 -------(1) Sum of products in pairs = 624
i.e. a $\times$ ar + ar $\times ar^{2}$ + a $\times ar^{2}$ = 624
⇒ $a^{2}r (1 + r + r^{2}$) = 624 -----(2)
Divide equation (2) by (1) we get,
ar = 12 ⇒ a = $\frac{12}{r}$
Put a = $\frac{12}{r}$ in equation (1) we get
$\frac{12}{r}(( 1 + r + r^{2}$)= 52

⇒ 3$r^{2}$ - 10r + 3 = 0
Factorizing
(3r -1)(r -3) = 0
∴ r = $\frac{1}{3}$ or r = 3
If r = 3 then a = 4 and if r = $\frac{1}{3}$ then a = 36
∴ three terms of G.P are 4,12,36 or 36,12,4.

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