# Geometric Series

Geometric Series form a very important section of the IBPS PO, SO, SBI Clerk and SO exams. A geometric series is also known as the geometric progression. It is a series formed by multiplying the first term by a number to get the second term, this process is continued until we get a number series in which each number is some multiple of the previous term. Such a progression increases swiftly and thus has the name geometric progression.

## Geometric Series or Geometric Progression

Geometric Progression or a G.P. is formed by multiplying each number or member of a series by the same number. This number is called the constant ratio. In a G.P. the ratio of any two consecutive numbers is the same number that we call the constant ratio. It is usually denoted by the letter ‘r’. Thus if we have a G.P. say a1, a2, a3, …, an, the ratio of any two consecutive numbers within the series will be same. Therefore for the series present above, we shall have:

a3/a2 = r; where ‘r’ is the common ratio. In other words, if you know ‘r’ and the first term, you can generate the entire Geometric Progression.

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Example 1: In a G.P., r = 2 and a = 1. Then the tenth term of the G.P. will be?

A) 16            B) 19           C) 26        D) 512

Answer: In a G.P. as we saw, each term is multiplied by the common ration ‘r’. To get the second term, the first term is multiplied by ‘r’. We get the third term by multiplying the first term by ‘r2‘. Similarly, we will get the fourth term by multiplying the first term by r3 and so on. Knowing this the above example becomes very easy. Since the first term is 1, we have to multiply it by 29 to get the tenth term = 512. So the tenth term of the G.P. = 512. The correct option thus is D) 512.

### Sum of the Geometric Progression

Sometimes you will be given the series and asked to find the sum of the first few terms or the entire series. The sum is denoted by Sn; where ‘n’ is the number of the term up to which the sum is being found out. For example, the sum of the first ten terms will be denoted by S10. Here we will list important formulae to find out the sum of the first few terms. Let ‘a’ be the first term of a G.P. and ‘r’ be the common ratio, then the sum of the G.P. can be found out by the following formulae:

S= a (r-1)/ r-1,                    if r ≠1  and

S= an ,                                     if r = 1

Sum of infinite terms  of a G.P. in case of  -1 < r <1 is given by the following formula:

Sn= a/(1-r).

So there are three formulae depending on the value of ‘r’. We will see examples of each below.

## Formula One Sn = a (rn -1)/ (r-1)

This formula is only valid when r ≠1. For example, consider the following series.

Example 2: Find the sum of the first 5 terms of the following series. Given that the series is finite: 3, 6, 12, …

A) 92           B) 24         C) 93/4         D) 27

Answer: The first step is to confirm that the series is actually a G.P. You can verify it by dividing the consecutive terms. Remember divide two sets of consecutive terms. For example, in the above example, 6/3 = 2 and 12/6 = 2. Hence the series is a G.P. with a common ratio or r = 2. Also, we see that a = 3, thus we can use the first formula and find the sum of any number of terms of such series.

To find the sum of the first 5 terms, we note that n = 5, a = 3, and r = 2. Thus we have:

S5 = 3(25 – 1)/(5 -1) = 93/4

Thus the option is C) 93/4.

## Second And Third Formulae For The Sum

The second formula works only when r = 1. this is pretty straightforward. In this case, each term of the G.P. will be same. The following trick question may be asked from this concept.

Q 1: What type of series is the following sequence of ‘n’ numbers:

1, 1, 1, 1, 1, 1, 1, ….., 1

B) Simple Series

D) Geometric Progression

Answer: The above series is clearly a Geometric Progression with the first term = 1 and the common ratio or r = 1 also. The sum of ‘n’ terms will be n(1) = n. Therefore, the correct option is D) Geometric Series.

The third formula is only applicable when the number of terms in the G.P. is infinite or in other words, the series doesn’t end anywhere. Also, the value of r should be between -1 and 1 but not equal to any of the two. -1 < r <1.

## Practice Questions

Q 1: In a Geometric Progression, the first term a = 10. The ratio of two consecutive terms is 2. What will be the sixth term of the series?

A) 320         B) 640           C) 300          D) 298

Ans: A) 320.

Q 2: Find the sum of the first five terms of the G.P.: 5, 25, 125, 625,… The G.P. has a thousand terms in it.

A) 395         B) 3905         C) 935         D) 9305

Ans: B) 3905

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##### Number Series

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Guest

5 7 31 283 ?

Guest
Parth Joshi

4533

Guest

3967

Guest
Priya

5×1+2=7
7×4+3=31
31×9+4=283
283×16+5=4533

Guest

ook

Guest
Venu

4533

Guest
anmol

4,8,24,28,84,88,_

Guest
Parth Joshi

264

Guest
Rahul

4+4=8×3=24+4=28×3=84+4=88×3=264

Guest
Ashish

84+100=184or 84+240=324

Guest
spy7

4+4=8
8×3=24
24+4=28
28×3=84
84+4=88
88×3=264

Guest
pavan kalyan

1 5 20 ???

Guest
lavanya lakhotia

60

Guest
anmol

1*3+2=5
5*3+5=20
20*3+7=67
can it not be like this???
if not then why?

Guest

1×5=5
5×4=20
20×3=60

Guest
Saumit

How

Guest
srashti

60

Guest
anmol

1*3+2=5
5*3+5=20
20*3+7=67
can it not be like this???
if not then why?

Guest
anmol

thats 20*3+8=68

Guest
Ashish

1×2+3=5×3+5=20×4+7=86

Guest
Honey

16,4,68,12,?,4,30,1,9 plz reply i need it

Guest
ABCD

GAY

Guest
Ayushi shukla

2,3,3,5,10,13,?,43,172,177

Guest
Sneha

4,5,5,7,9,13,10__,14

Guest
abcd

I think
In series of odd numbers (4, 5,9,10,14) there is addition of 1 and 4 alternately.
And in series of even numbers (5, 7,13,?) There is addition of 2 and 6 alternately.

Guest
Janu

Find missing teams-1,5,14,?,44

Guest
Robin rathi

Find the missing number of this series
60,50,60,90,41,_?

Options 1. 12
2. 18
3. 25
4. 30
5. none this above

Guest
Rohith

26,4,20,10,14,16,8,22,2,28

Guest

____,360,000,000,____, 389,000,000,____, 420,000,000

Guest
Sune Pedersen

37 52 93 75 29 ? what is the math behind this

Guest
Sonali sahu

2,1,0,-3,-24,? Find the next number

Guest

Upper line 3 5 8 mid line 6 10 32 lower line 9 ? 50 me missing no. Kya h

Guest
Akash

94 101 115 136 164 ?

Guest
Ashish

199

Guest
Lokesh

QID : 426 – In the following question, select the
missing number from the given alternatives.
41, 83, 167, 335, 671, ?
Options:
1) 1297
2) 1343
3) 1447
4) 1661

Guest
Ashish

1343

Guest
Shriya Nimje

50,50,54,72,?,220