JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Sequences and Series

Let the positive numbers

a_{1}, a_{2}, a_{3}, a_{4}

and

a_{5}

be in a G.P. Let their mean and variance be

\frac{31}{10}

and

\frac{m}{n}

respectively, where m and n are co-prime. If the mean of their reciprocals is

\frac{31}{40}

and

a_{3} + a_{4} + a_{5} = 14

, then

m + n

is equal to ______.

Answer: 211

Solution

Let the five terms in G.P. be

\frac{a}{r ^{2}}, \frac{a}{r}, a, a r, a r^{2}

.

Given mean =

\frac{31}{10}

\frac{a}{r ^{2}} + \frac{a}{r} + a + a r + a r^{2} = 5 \times \frac{31}{10} = \frac{31}{2}

... (1)

Mean of reciprocals =

\frac{31}{40}

\frac{r ^{2}}{a} + \frac{r}{a} + \frac{1}{a} + \frac{1}{a r} + \frac{1}{a r ^{2}} = 5 \times \frac{31}{40} = \frac{31}{8}

... (2)

Dividing (1) by (2):

a^{2} = 4

, so

a = 2

(positive).

Substituting in (1):

r + \frac{1}{r} = \frac{5}{2}

, which gives

r = 2

r = \frac{1}{2}

.

Taking

r = 2

, the G.P. is:

\frac{1}{2}, 1, 2, 4, 8

.

Variance

σ^{2} = \frac{\sum x _{i}^{2}}{5} - (\frac{31}{10})^{2} = \frac{\frac{1}{4} + 1 + 4 + 16 + 64}{5} - \frac{961}{100} = \frac{341}{20} - \frac{961}{100} = \frac{1705 - 961}{100} = \frac{744}{100} = \frac{186}{25}

.

Since

g cd (186, 25) = 1

, we have

m = 186, n = 25

.

Therefore

m + n = 211

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer