Question

A lucky man finds 6 pots of gold coins .He counts the coins in the 1st four pots to be 60,30,20,15 respectively. If there is a definite progression , what would be numbers of coins in the net two pots?

Answer 1

Given:

Total number of pots = 6

Coins in first four pots = 60,30,20 and 15

To Find:

The coins in other two pots

Solution:

Coins in first pot = 60 × 1/2 = 30

Coins in second pot = 30 × 2/3 = 20

Coins in third pot = 20 × 3/4 = 15

Now,

15 × 4/5 = 12 and

12 × 5/6 = 10

Answer: Thus, the number of coins in next two pots is 12 and 10

Answer 2

Given :

Total no of pots containing gold coins = 6

No of coins in the 1st four pots = 60 , 30 , 20 , 15 respectively

To Find :

No of coins in the next two pots = ?

Solution :

From the given sequence for the number of coins in the 1st four pots we can observe that the given sequence form a GP  .

We know that for a GP  , reciprocal of the terms forma an AP .

So , $\frac{1}{60} , \frac{1}{30}, \frac{1}{20}, \frac{1}{15}$ forms an AP .

And  common difference for this AP will be :

= $\frac{1}{30}- \frac{1}{60} = \frac{12-1}{60}= \frac{1}{60}$

So , the 5th term of this AP will be :

= a + (n-1)d

= $\frac{1}{60} + (5-1) \times \frac{1}{60}$

= $\frac{1}{60} +\frac{4}{60}$

=$\frac{5}{60}$

= $\frac{1}{12}$

Since  $\frac{1}{12}$ is 5th term of the AP , so the 5th term of the GP will be reciprocal of this :

=$\frac{1}{\frac{1}{12}}$

= 12

And the 6th term of the AP will be :

= a + (n-1)d

= $\frac{1}{60} + (6-1) \frac{1}{60}$

= $\frac{1}{60} + 5 \times \frac{1}{60}$

= $\frac{5+1}{60}$

=  $\frac{6}{60}$

= $\frac{1}{10}$

Now again  $\frac{1}{10}$ is the 6th term of the AP, so the 6th term of the GP will be reciprocal of this :

= 10

Hence , number of coins in 5th pot is 12 and in 6th pot 10 .