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Sequence and Series

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Question : 16 of 86

Marks: +1, -0

Find the three numbers in G.P whose sum is 52 and the sum of their product in pairs is

624 .

$4,12,36$
$10,16,26$
$5,17,30$
None

Solution:

Concept:

Let three numbers in G.P. be

a, ar, ar^2

. Use sum and sum of pairwise products to find

a

and

r

.

Explanation:

Sum:

a + ar + ar^2 = a(1+r+r^2) = 52

Sum of products in pairs:

a \cdot ar + ar \cdot ar^2 + a \cdot ar^2 = a^2r + a^2r^3 + a^2r^2 = a^2r(1+r+r^2) = 624

From first,

1+r+r^2 = \frac{52}{a}

. Substitute into second:

a^2r \cdot \frac{52}{a} = 52ar = 624

So

ar = 12

(the middle term).

Then

a + ar^2 = 52 - 12 = 40

. Since

a = \frac{12}{r}

, we have

\frac{12}{r} + 12r = 40

Multiply by

r

:

12 + 12r^2 = 40r \Rightarrow 12r^2 - 40r + 12 = 0

Divide by 4:

3r^2 - 10r + 3 = 0 \Rightarrow (3r-1)(r-3)=0

So

r = 3

or

r = \frac{1}{3}

. For

r=3

,

a = \frac{12}{3}=4

, numbers:

4, 12, 36

.

For

r=\frac{1}{3}

,

a=36

, same set.

Answer:

The numbers are

4, 12, 36

, which corresponds to option A.

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