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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 16-18

Direction: Consider the following for the next three (03) items that follow:

Let

a, x, y, z, b

be in AP, where

x + y + z = 15

. Let

a, p, q, r, b

be in HP, where

p^{-1} + q^{-1} + r^{-1} = \frac{5}{3}

Question : 18 of 120

Marks: +1, -0

What is the value of pqr?

Solution:

Concept:

If five reciprocals are in arithmetic progression (AP), the common difference is found using the last term formula.

Explanation:

Given

\frac{1}{a}, \frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{b}

are in AP.

Take

a = 9

and

b = 1

(from the previous part of the problem).

Using

n

th term of AP:

\frac{1}{b} = \frac{1}{a} + (5-1)d

Substitute values:

1^{-1} = 9^{-1} + 4d

1 = \frac{1}{9} + 4d

⇒

4d = \frac{8}{9}

⇒

d = \frac{2}{9}

Now compute each reciprocal:

\frac{1}{p} = \frac{1}{a} + d = \frac{1}{9} + \frac{2}{9} = \frac{1}{3}

⇒

p = 3

\frac{1}{q} = \frac{1}{a} + 2d = \frac{1}{9} + \frac{4}{9} = \frac{5}{9}

⇒

q = \frac{9}{5}

\frac{1}{r} = \frac{1}{a} + 3d = \frac{1}{9} + \frac{6}{9} = \frac{7}{9}

⇒

r = \frac{9}{7}

Multiply:

pqr = 3 \times \frac{9}{5} \times \frac{9}{7} = \frac{243}{35}

Answer:

\frac{243}{35}

(Option C).

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