Question

Find five numbers in g.P. Such that their product is 1024 and fifth term is square of third term

Answer 1

Answer:

The five numbers are

1, 2, 4, 8, 16

or

1, -2, 4, -8, 16

Step-by-step explanation:

Let the first term of the GP be a and the common ratio be r

Then the five terms are

$a, ar, ar^2, ar^3, ar^4$

Product of the terms

$P=a\times ar\times ar^2\times ar^3\times ar^4$

$\implies P=a^5r^{10}$

$\implies P=(ar^2)^5$

Given the product P = 1024

Therefore,

$(ar^2)^5=1024$

or, $(ar^2)^5=2^{10}$

$\implies ar^2=2^{2}=4$  ...... (1)

but $ar^2$ is the third term of the GP

Also given fifth term is the square of the third term

Therefore

$ar^4=(ar^2)^2=4^2=16$

Dividing the fifth term by 3rd term we get

r^2=4

or, r = 2, -2

Therefore from (1)

a = 1

Thus two GPs are possible

1, 2, 4, 8, 16

1, -2, 4, -8, 16

Answer 2

Answer:

Required G.P :

1,2,4,8 and 16

Or

1,-2, 4, -8 and -16

Step-by-step explanation:

$Let \: \frac{a}{r^{2}},\frac{a}{r},a,ar,ar^{2}\: are \\ first \:five \: terms \: of \: G.P$

$Product \: five \: terms = 1024$

$\implies \frac{a}{r^{2}}\times \frac{a}{r}\times a\times ar\times ar^{2} = 4^{5}$

$\implies a^{5} = 4^{5}$

$\implies a = 4$

/* According to the problem given,

$fifth \:term\: is\: square\: of\: third \:term$

$\implies ar^{2}=a^{2}$

$\implies r^{2}= a$

$\implies r^{2}= 4$

$\implies r = ±\sqrt{4}=±2$

Now,

Required G P :

Case 1 :

If a = 4 , r = 2

$\frac{4}{2^{2}},\frac{4}{2},4,4\times 2,4\times 2^{2}$

1,2,4,8 and 16

Case 2 :

If a = 4 , r = -2, then

G.P :

1 , -2, 4, -8 and 16

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