Geometric progression problems and solutions with Formulas and properties

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.

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Formulas and properties of Geometric progression

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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, ar², ar³. . . . . . .

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 / 2ⁿ

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

a₁₅ = [ 192 / (2)⁷ ] (2) ¹⁴

a₁₅ = 192 x 2 ⁷= 3 x 2 ⁶x 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 a₄ = 27 and let common ratio is ‘r’ . So

⇒ a₄ = a r^4-1

⇒ 27 = 1 r^4-1 = r³

⇒ Common ratio = r = 3

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

Solution: If a₁, a₂, a₃ , . . . . . . is a GP then

i.e Common ratio = r = a₂ / a₁= a₃ / a₂= a₄ / a₃= . . . . .

So ⇒ $\frac{a+2}{a} \ = \frac{a+6}{a+2} \$

⇒ a² + 6a = a² + 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 $\frac{b-c}{a-b} \ = \frac{c-a}{b-c}$

⇒ b² + c² – 2bc = ac – a² -bc + ba

⇒ a² + b² + c² = ab + bc + ca —————– ( i )

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

(a +b + c)² = a² + b² + c² + 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 a₆ = 24 ————- ( i) & a₁₃ = 3/16 ————– ( ii)

⇒ a₆ = a r^6-1& a₁₃ = a r^13-1

⇒ 24 = a r⁵& 3/16 = a r¹²

⇒ r⁷= 3 / 24 x 16 = 1 / (2)⁷

⇒ r = 1/2 ———– (iii)

from the above equations

⇒ a₆ = 24 = a (1/2)⁵

⇒ a = 3 x 2⁸

Now a₂₀ = a r^20-1

a₂₀ = 3 x 2⁸ x ( 1/2 )¹⁹ = 3 / 2¹¹

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 a₇ = 8 x a₄and also a₅ = 48

⇒ a r^7-1 = 8 x a r^4-1

⇒ r⁶ = 8 x r³

⇒ r = 2

Now take a₅ = 48

⇒ a r^5-1= 48

⇒ a 2⁴= 48

⇒ a = 3

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

Solution: Formula – The ‘n’ numbers G₁, G₂, G₃, . . . . . . . G_n are said to be Geometric means in between ‘a’ and ‘b’. If a, G₁, G₂, G₃, . . . . . . . G_n, b are in G.P.

So second geometric mean in sequence is $G_{2} \ = a \ ( \frac{b}{a} ) ^{ \frac{2}{n+1}}$

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

G₂ = (1/8) ( 128 x 8)^2/5

G₂ = 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 = $\sqrt{xy}$

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

a₅ = 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 cm²

Second square = 16² + 16² = 512 cm² ( 1024/2 = 512)

Third square = 16² = 256 cm² ( 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 cm²

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