nth term of a Geometric Progression

Let us examine a problem. The number of bacteria in a certain culture triples every hour. If there were 30 bacteria present in the culture originally, then, what would be the number of bacteria in the fourth hour?

To answer this let us first see what the number of bacteria in the second hour would be.

Since for every hour it triples

No. of bacteria in the second hour = 3 × no. of bacteria in the first hour

= 3 × 30 = 30 × 31

= 30 × 32−1

=

No. of bacteria in the third hour = 3 × no. of bacteria in the second hour

= 3 × 90 = 30 × (3 × 3)

= 30 × 32 = 30 × 33−1

=

No. of bacteria in the fourth hour = 3 × no. of bacteria in the third hour

= 3 × 270 = 30 × (3 × 3 × 3)

= 30 × 33 = 30 × 34−1

=

Observe that we are getting a list of numbers

30, 90, 270, 810, ....

These numbers are in GP (why?)

Now looking at the pattern formed above, can you find the number of bacteria in the 20th hour?

You may have already got some idea from the way we have obtained the number of bacteria as above.

By using the same pattern, we can compute that the number of bacteria in the 20th hour.

= 30 × (3 × 3 × ... × 3)

= 30 × 319

= 30 × 320−1

This example would have given you some idea about how to write the 25th term, 35th term and more generally the nth term of the GP.

Let a1, a2, a3, .... be in GP whose 'first term' a1 is a and the common ratio is r.

then the second term

a2 = ar = ar2−1

the third term

a3 = a2 × r = (ar) × r = ar2 = ar3−1

the fourth term

a4 = a3 × r = ar2 × r = ar3 = ar4−1

. . . . . . . . . . . . . .

. . . . . . . . . . . . . .

Looking at the pattern we can say that nth term an = arn−1

So nth term of a GP with first term 'a' and common ratio 'r' is given by an = arn−1.