Geometric progression problems and solutions with Formulas and properties

This page teaches you about the Geometric Progression Tutorial—the nth term of GP, the sum of GP, and geometric progression problems with solutions for all competitive exams and academic classes.

Geometric Sequences Practice Problems | Geometric Progression Tutorial

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, . . . . . . .

The 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

The 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 the 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

⇒ common $\frac{b-c}{a-b} = \frac{c-a}{b-c}$

⇒ 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 the first term is ‘a’ and the 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 the fourth term. What will be the first term when its 5^th is 48?

Solution: Let the first term be ‘a’ and the 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 the 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 the second geometric mean in a sequence is $G_{2} = a \left( \frac{b}{a} \right)^{ \frac{2}{n+1} }$

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 is deposited in the bank with an annual interest rate 10% interest compounded annually, then find the total amount at the end of the first, second, third, fourth and first years.

Solution: Here a = 1000 (starting of the first year) and common ratio = 1.1

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

At the start 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 a yearly increment of 10%. Now find the Robin’s salary at the start of 5^th year.

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

Robin’s salary at the start of 5^th year i.e n = 5

The 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 drawn by joining the midpoints of the sides of a given square. A second square is drawn inside the second square in the same way the second square is drawn inside the third square, and this process continues up to 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, the areas of the squares are in geometric progression

so the remaining areas of squares are 128, 64, 32, 16 and 8 cm²

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