Geometric Progression

The geometric progression is a sub-topic of progression and series and it holds a good weightage in the CAT quantitative section. A geometric progression “GP” consists of a sequence of numbers such that the value obtained by the division of the (n+1)th term by the nthterm is the same as that obtained when the nth term divided by the (n – 1)th term.

This common value is called the common ratio(r).

For example, take a sequence of numbers as: 2, 4, 8, 16 Now 4/2=8/4=16/8= 2. So this sequence of numbers forms a GP with a common ratio = 2.

General form of a GP

Consider a GP with first term as ‘a’ and common ratio as ‘r’.

1st term = a

2nd term = ar

3rd term = ar2

4th term = ar3

Proceeding this way,

nth term = Tn= arn-1

Sum of GP: Formula

The nth term of a geometric progression is given by Tn=arn-1,

So sum to ‘n’ terms,

Sn = a + ar + ar2 + ar3 + … + arn-1

= a (1 + r + r2 + r3 + … + rn-1)

= a ((1- rn ))/((1-r) ) if r < 1 & a ((rn-1))/((r-1) ) if r > 1

The sum of geometric series up to n terms is given by,

Sn= a(1- rn )/((1-r) ) (1 > r)

S= a ((rn-1))/((r-1) ) (1 < r)

Geometric Progression Questions

Question: Find the eight term of the series: 1/3 – 1/6 + 1/12 – 1/24

a) 1/384

b) 1/192

c) (-1)/192

d) (-1)/384

e) (-1)/128

Solution:

The series is a GP with common ratio r = (((-1)/6)/(1/3)) = ((-1)/2), Eight term = ar7 = (1/3) × ((-1)/2)7 = (-1)/384

Hence option (d)

Question: Find the sum of the first five terms of the series: 3, 12, 48…

a) 512

b) 1024

c) 198

d) 343

e) 128

Solution:

Sn=[a(rn– 1) ]/(r – 1)

In the question a = 3, r = 4 & n = 5

S5= 3 ×(45– 1)/(4 – 1)= 1024

Hence Option (b)

Question: The sum of first two terms of a G.P. is 5/3 and the sum to infinite terms is 3. If the GP has a positive common ratio, what is the first term?

a) 1

b) 6

c) 9

d) 3/5

e) can’t be determined

Solution:

The first two terms = a, ar a + ar = 5/3

a/(1 – r)= 3 ⇒ a = 3(1-r)a(1 + r)=5/3

⇒3(1-r)(1 + r)=5/3⇒1 – r2 = 5/9 ?r2 = 4/9 ? r = 2/3

∴ a = 3(1-2/3)= 1
So, first term is 1.

Hence option (a)

Question: The sum of an infinite GP whose common ratio is positive and numerically less than 1 is 32 and the sum of first two terms is 24.

What will be the third term?

Solution:

S = a/(1 – r) for an infinite GP with r < 1

⇒32 = a/(1 – r) ? a = 32 (1 – r)

First term = a

Second term = ar

a + ar = 24 ? a = 24/(1 + r)

⇒32(1 – r) = 24/( 1 + r)

⇒1 – r2 = 24/32 = 3/4

⇒r2 = 1 – 3/4 = 1/4 ? r = 1/2

So a = 32(1/2) = 16

Third term = ar2 = (16)(1/4)= 4.

Question: Determine the first term of a GP, the sum of whose first term and third term is 40 and the sum of whose second term and fourth term is 80.

Solution:

Suppose the GP has first term as ‘a’ and common ratio as ‘r’

First term = a

Second term = ar

Third term = ar2

Fourth term = ar3

a + ar2 = 40 ? a(1 + r2) = 40 —(i)

ar + ar3 = 80 ? ar ( 1 + r2) = 80 —(ii)

dividing (ii) by (i)

r = 2

a = 40/((1+4) ) = 8

So, the first term = 8.

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