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As we saw arithmetic progressions in the previous article, we would be looking at another popular progression type in this article viz. the geometric progression.

What is a geometric progression?

A geometric progression is similar to an arithmetic progression in the manner that there is a series with a fixed relationship between consecutive terms. The fixed relationship here is the common ratio that exists between the two terms.

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Here, the common notation is ‘a’ for the first term, and ‘r’ for the common ratio. The nth term is Tn and the sum of the first n terms is given by S.

The questions primarily involve usage of one or more of the following

1) Value or form of the nth term of the geometric progression

2) Value or form of the common ratio of the geometric progression

3) Sum of terms in a geometric progression

4) Sum of infinite geometric progression

5) Applications of sum of infinite geometric progression

The formulas to remember and the important observations are:

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If a, b, c are in geometric progression, then b2 = ac

If Tm and Tn are known such that n > m, the common ratio can be easily found by

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If r > 1, the sum of all terms of an infinite geometric progression cannot be determined as it would tend towards infinity.

T1 * Tn = T2 * Tn-1 = T3 * Tn-2

If three terms are said to be in a geometric progression it would be beneficial to take the terms as follows especially if the product of the three numbers is given

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Let’s see a few questions on geometric progression and a few of its interesting applications.

The sum of first 6 terms of a geometric progression is equal to the sum of the first 8 terms of the same geometric progression. Sum of the first 9 terms is 28 and all the elements of the geometric progression are non-zero, what is the third term in the geometric progression?

Sum of the first 8 terms would include the first 6 terms and also the 7th and the 8th term. Now, we understand that the sum of the 7th and the 8th terms is 0.

So, ar6 + ar7 = 0

ar6(r + 1) = 0

Either a = 0 or r = 0 or r = –1

As the sum of the first 9 terms is 28, a cannot be equal to 0. Also, r cannot be equal to 0 as the terms in the geometric progression are non-zero in nature. So, r = –1

S9 = a, S8 ­= 0, S7 = a and so on. So, we can safely say that T1 = a, T2 = –a, T3 = a. As all odd numbered terms are equal to the first term and the sum of the first 2n terms is always equal to zero, T3 = a = 28.

What is the remainder when the sum of the series 5 + 55 + 555 + 5555 + … 99 terms is divided by 400? (Probable NMAT question)

In this type, you have to proceed as follows:

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which will give you 679 mod 8 – 55 mod 400

which will be 7*50 – 55 mod 400 = 295 mod 400 which is your answer.

What is the fractional representation of the number 0.145454545…?

N = 0.1454545…

10N = 1.454545…

1000N = 145.4545…

990N = 144

N = 144/990

The other way to do this is using the infinite geometric progression formula.

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A ball bounces from a height of 2 metres and returns to 80% of its previous height on each bounce. Find the total distance traveled by the ball until it stops bouncing.

This is again an interesting application of geometric progression.

You know that the ball has been dropped from a height of 2 m and then it bounces to 4/5th of the height from which it starts falling. So, the distance traveled would look as follows:

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Total of 18 m.

There is another method to do this as well. You simply need to know the height (h) from which the ball is dropped and the basic integral values of ‘a’ and ‘b’ where a/b is the percentage of the previous height that a ball reaches in the next bounce.

The total distance covered would be simply

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A square is drawn by joining the midpoints of the sides of a given square, a third square drawn inside the second square in the same way and this process continues indefinitely, if the side of the first square is 16 cm, what is the sum of areas of all the squares?

  1. a) 1024 b) 512 c) 256          d) Cannot be determined

Area of the first square would be 162 = 256 sq.cm

For the second square, the length of the side of the first square becomes the diagonal. So, length of diagonal of the second square is 16 cm and so, length of side of the second square is 8√2 cm.

Area of the second square would be (8√2)2 = 128 sq.cm

Area of the third square would be 82 = 64 sq.cm

So, this is nothing but the sum of an infinite geometric progression

S = 256 + 128 + 64 + …

S = 512 sq.cm

Strategy and final words

So overall, geometric progression is a fairly easy topic with questions being direct formula based. It takes a trained eye to spot the connections between the terms and figure out if it is an infinite geometric progression question or not. The key is to understand that the common ratio would be less than 1. The questions from geometric progression can broadly be divided into two types (a) the ones which involve the sum of infinite geometric progressions and (b) Direct application of the basic formulae in the form of pure geometric progression questions, compound interest based questions, mixtures and alligations based questions, sum of factors of a number, etc. The former is easy to solve and one needs to simplify the data in terms of fractions without actually calculating them whereas the latter involves a bit of calculation and sometimes a bit of vision to understand which number/s would fit into the expressions. Personally, if it is an infinite geometric question, it should count as a must attempt in the first round but if it involves pure calculation, it would probably be a good idea to attempt it at a later stage while solving the paper.

Hope the article cleared a few doubts regarding geometric progressions. All the best!

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