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}

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