In this page learn about *Geometric Progression Tutorial* – n^{th} term of GP, sum of GP and *geometric progression problems with solution* for all competitive exams as well as academic classes.

## Geometric Sequences Practice Problems | Geometric Progression Tutorial

**Formulas and properties of Geometric progression**

**Geometric progression exercises with answers**

Example – 1 : Write G.P if a = 128 and common ratio r = -1/2

Solution : General form of G.P = a, ar, ar^{2}, ar^{3 }. . . . . . .

128, -64, 32, -16, . . . . . . .

Formula for n^{th} term G.P is a_{n} = ar^{n-1}

Example- 2: Find the 10^{th} and n^{th} term of the Geometric sequence 7/2, 7/4, 7/8, 7/16, . . . . . . .

Solution: Here a = 7/2 and common ratio = (7/4) / (7/2) = 1/2

Formula for n^{th} term G.P is a_{n} = ar^{n-1}

10^{th} term of given G.P = (7/2) x (1/2)^{9}= (7/2) x (1/514) = 7/1024

n^{th} term = (7/2) x (1/2)^{n-1 }= 7 / 2^{n}

Example- 3 The number 2048 is which term in the following Geometric sequence 2, 8, 32, 128, . . . . . . . . .

Solution: Here a = 2 and r = 4

n^{th} term G.P is a_{n} = ar^{n-1}

⇒ 2048 = 2 x ( 4) ^{n-1}

⇒ 1024 =( 4) ^{n-1}

⇒ ( 4) ^{5} = ( 4) ^{n-1}

⇒ n = 6

Note: Given G.P is simple numbers. So we can find easily by direct multiplication upto required number.

Example – 4: Find the 15^{th} term of a G.P Whose 8^{th} term is 192 and the common ratio is ‘2’

Solution: Let first term of G.P is ‘a’ and common ratio r = 2

8^{th} term of G.P is 192 So

⇒ 192 = a x (2)^{7}

⇒ a = 192 / (2)^{7}

Now 15^{th} is

a_{15 } = [ 192 / (2)^{7} ] (2) ^{14}

a_{15 } = 192 x 2 ^{7 }= 3 x 2 ^{6 }x 2 ^{7 }= 3 x 2 ^{13}

Example – 5: In a G.P first term is ‘1’ and 4^{th} term is ‘ 27’ then find the common ration of the same.

Solution: Here a = 1 and a_{4} = 27 and let common ratio is ‘r’ . So

⇒ a_{4} = a r^{4-1}

⇒ 27 = 1 r^{4-1} = r^{3}

⇒ Common ratio = r = 3

Example – 6: Find ‘a’ so that a, a+2, a+6 are consecutive terms of a geometric progression.

Solution: If a_{1}, a_{2}, a_{3} , . . . . . . is a GP then

i.e Common ratio = r = a_{2} / a_{1 }= a_{3} / a_{2 }= a_{4} / a_{3 }= . . . . .

So ⇒

⇒ a^{2} + 6a = a^{2} + 4 + 4a

⇒ a = 2

Example – 7 : If (a-b), (b-c), (c-a) are the consecutive terms of G.P then find (a +b + c)^{2}

Solution: As per properties of Geometric Progression

⇒ commoon

⇒ b^{2} + c^{2} – 2bc = ac – a^{2} -bc + ba

⇒ a^{2} + b^{2} + c^{2} = ab + bc + ca —————– ( i )

Now take (a +b + c)^{2} as per algebraic formula

(a +b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca) —————– ( ii )

From the above two equations ( i ) & ( ii )

(a +b + c)^{2} = 3 (ab + bc + ca)

Example – 8: In a G.P 6^{th} term is 24 and 13^{th} term is 3/16 then find 20^{th} term of the sequence.

Solution: Let first term is ‘a’ and common ratio is ‘r’

Here a_{6} = 24 ————- ( i) & a_{13} = 3/16 ————– ( ii)

⇒ a_{6} = a r^{6-1 }& a_{13} = a r^{13-1 }

⇒ 24 = a r^{5 }& 3/16 = a r^{12}

⇒ r^{7 }= 3 / 24 x 16 = 1 / (2)^{7}

⇒ r = 1/2 ———– (iii)

from the above equations

⇒ a_{6} = 24 = a (1/2)^{5 }

⇒ a = 3 x 2^{8}

Now a_{20 } = a r^{20-1}

a_{20 } = 3 x 2^{8} x ( 1/2 )^{19} = 3 / 2^{11}

Example – 9: The 7^{th} term of a G.P is eight times of fourth term. What will be the first term when its 5^{th} is 48?

Solution:Let first term is ‘a’ and common ratio is ‘r’ for the given geometric sequence

Here given a_{7} = 8 x a_{4 }and also a_{5} = 48

⇒ a r^{7-1} = 8 x a r^{4-1}_{
}

⇒ r^{6} = 8 x r^{3}

⇒ r = 2

Now take a_{5} = 48

⇒ a r^{5-1 }= 48

⇒ a 2^{4 }= 48

⇒ a = 3

Example -11: Four geometric means are inserted between 1/8 and 128. Find second geometric mean.

Solution: Formula – The ‘n’ numbers G_{1}, G_{2}, G_{3}, . . . . . . . G_{n} are said to be Geometric means in between ‘a’ and ‘b’. If a, G_{1}, G_{2}, G_{3}, . . . . . . . G_{n}, b are in G.P.

So second geometric mean in sequence is

In the given sum a = 1/8 , n = 4 and b = 128

G_{2} = (1/8) ( 128 x 8)^{2/5}

G_{2} = 2

Example-12: ‘x’ and ‘y’ are two numbers whose AM is 25 and GM is 7. Find the numbers

Solution: Here x’ and ‘y’ are two numbers then

Arithmetic mean = AM = (x+y)/2

Geometric Mean =GM =

So

x + y = 50 ————— ( i) & xy = 49 ———- (ii)

From the above equations the number can be 1 & 49

**Application of geometric progression**

Example – 1: If an amount ₹ 1000 deposited in the bank with annual interest rate 10% interest compounded annually, then find total amount at the end of first, second, third, forth and first years.

Solution: Here a = 1000 (staring of first year) and common ration = 1.1

So the total amount at the end of first, second, third, forth and first years will be in geometric sequence

At staring of 1^{st} year – 1000

For the end of the 1^{st} year – 1000 x 1.1 = 1100

For the end of the 2^{nd} year – 1100 x 1.1 = 1210

For the end of the 3^{rd} year – 1210 x 1.1 = 1331

For the end of the 4^{th} year -1331 x 1.1 = 1464.1

For the end of the 5^{th} year – 1464.1 x 1.1 = 1610.5

Example – 2: Salary of Robin, When his salary is ₹ 5,00,000 per annum for the first year and expected to receive yearly increment of 10%. Now find the Robin salary at staring of 5^{th} year.

Solution: Here a = 5,00,000 and common ratio = 1.1

Robin salary at staring of 5^{th} year i.e n = 5

Formula for n^{th} term G.P is a_{n} = ar^{n-1}

a_{5} = 500000 x (1.1)^{4 }= 500000 x 1.4641 = 7,32,050

Example – 3: The square is draw by joining the mid points of the sides of given square. A second square is drawn inside the second square in the same way second square is drawn inside the third square and this process continues upto 8^{th} square. If a side of the first square is 32 cm then find the area of the first to eight square

Solution: Here area of the first square = 32 x 32 = 1024 cm^{2}

Second square = 16^{2} + 16^{2} = 512 cm^{2} ( 1024/2 = 512)

Third square = 16^{2} = 256 cm^{2} ( 512/2 = 256)

From the above, areas of the squares are in geometric progression

so remaining areas of squares are 128, 64, 32, 16 and 8 cm^{2}

In the same way remaining areas of squares are 128 cm^{2}, 64 cm^{2}, 32 cm^{2}, 16 cm^{2} and 8 cm^{2}

**Related Articles**

Arithmetic Progression Formulas

Arithmetic Progression Questions with Solutions

Arithmetic Mean formula with Examples

Geometric Progression formulas

Geometric Progression Examples

Harmonic Progression formulas and examples

Relation Between arithmetic ,Geometric and Harmonic means

## 6 thoughts on “Geometric progression problems and solutions with Formulas and properties”

## Francis

(August 5, 2021 - 8:00 pm)Very good site

## sivaalluri

(August 7, 2021 - 5:31 am)Thanks

## Ajayi Dele

(October 24, 2021 - 2:02 pm)Pls I don’t really know maths

## Tony

(January 27, 2023 - 7:21 am)Please I really don’t know maths 🙄 help me

## Oluyole

(October 30, 2021 - 11:11 am)I need alert from this site

## Bashir Ibrahim

(June 9, 2022 - 9:00 am)Let my apologies for the solution of geometric progression