Geometric Progression Solved Problems

A geometric progression is a sequence where the succeeding term is ‘r’ times the preceding term. If a paper is folded four or five times, calculating the height of the stack of the paper after being folded four to five times is an example of geometric progression. In this article, we come across geometric progression solved problems along with the definition and properties of common ratio.

Geometric Progression Definition

A sequence in which the first term is non-zero and the ratio of each term to its previous term remains constant. The ratio is represented by the letter “r” and is called the common ratio.

The properties of a geometric progression are listed below:

• If a constant non-zero quantity is multiplied or divided by each term, the resulting sequence is also a geometric progression.
• The reciprocal of all the terms in geometric progression also forms a geometric progression.
• If all the terms in a geometric progression are raised to the same power, then the new sequence obtained is also in geometric progression.

Finite and infinite geometric progression

The geometric progression with a finite number of terms is a finite geometric progression, whereas a GP with an infinite number of terms is an infinite geometric progression.

Important Results

The formulae related to a geometric progression are given below:

To Determine Whether the Given Sequence Is a GP or Not

If the common ratio between the terms is a constant, then it is said to be a geometric sequence.

For instance, check whether the sequence 8, 16, 32, 64, 128 . . .  . . is a GP.

Solution: 

Given

Since the common ratio is a constant, it is a GP.

Properties of Common Ratio [r]

• The value of the common ratio determines whether the sequence is increasing or decreasing.
• r > 1 indicates that the GP has exponential growth towards positive infinity.
• r < -1 indicates that the GP has exponential growth towards infinity in the alternating sign.
• If the value of r lies between -1 and 1, then GP has exponential decay towards 0.

Related articles

Geometric Progression

Geometric Mean

Solve Problems on Geometric Progression for IIT JEE

Example 1: If the 4^th, 7^th and 10^th terms of a G.P. be a, b, and c, respectively, then what is the relation between a, b, and c?

Solution:

Let the first term of G.P. = A and common ratio = r

We know that the n^th term of G.P. = Arⁿ⁻¹

Now t₄ = a = Ar³, t₇ = b = Ar⁶ and t₁₀= c = Ar⁹

b² = (Ar⁶)²

= A²r¹²

ac =  (Ar³) (Ar⁹) = A²r¹²

So, b² = ac is the relation between a, b and c.

Example 2: If a, b, c are p^th, q^th and r^th terms of a G.P., then what is (c / b)^p (b / a)^r (a / c)^q equal to?

Solution:

a = AR^p−1, b = AR^q−1, c = AR^r−1

(c / b)^p * (b / a)^r * (a / c)^q = [(AR^r−1 / AR^q−1) ]^p * [(AR^q−1 / AR^p−1)]^r * [(AR^p−1/ AR^r−1)]^q

= R^{(r−q)p+(q−p)r+(p−r)q}

= R⁰

= 1

Example 3: If the 6^th term of a G.P. is 32, and its 8^th term is 128, then find the common ratio of the G.P.

Solution:

T₆ = 32 and T₈ = 128

⇒ ar⁵= 32 ….. (i) and

ar⁷= 128 …..(ii)

Dividing (ii) by (i), we have

r²=4

r = 2

Example 4: Find the sum of the series 6 + 66 + 666 + ……….upto n terms.

Solution:

Given series 6 + 66 + 666 + ……….upto n terms

= [6 / 9] (9 + 99 + 999 + ….. upto n terms)

= [2/3](10 – 1 + 100-1 + 1000-1 + …upto n terms)

= [2 / 3] ((10 + 10² + 10³ + ………. + upto n terms) −n)

= [2 / 3] [10 (10ⁿ−1) / 10−1] −n)

= [2 / 3] [10 (10ⁿ−1) / 9] −n)

= [2/3] [10ⁿ⁺¹ – 10 – 9n]/9

= 2 (10ⁿ⁺¹− 9n − 10) / 27

Example 5: The sum of a few terms of any ratio series is 728. If the common ratio is 3 and the last term is 486, then what will be the first term of the series?

Solution:

n^th term of series = arⁿ⁻¹ = a(3)ⁿ⁻¹ = 486 …. (i)

Sum of n terms of series is S_n= a (3ⁿ−1) / [3−1] = 728 ….(ii)(∵r >1)

From (i), a (3ⁿ / 3) = 486 or [a * 3ⁿ] = 3 × 486 = 1458

From (ii), [a * 3ⁿ] − a = 728 * 2 or [a * 3ⁿ] − a = 1456

1458 − a = 1456

or a = 2

Example 6: If three geometric means be inserted between 2 and 32, then find the third geometric mean.

Solution:

The GP will be 2, g₁ , g₂, g₃, 32 where a = 2, ar = g₁, ar² = g₂, ar³ = g₃ and ar⁴ = 32

Now, 2 × r⁴ = 32

⇒ r⁴ = 16 = (2)⁴

⇒ r = 2

Then, the third geometric mean = ar³ = 2 × 2³ = 16

Example 7: An increasing GP is formed by three positive numbers. The new numbers form an AP if the middle term of the geometric progression is doubled. Find the common ratio of the GP.

Solution: 

Let a, ar, ar² be in GP.

If the middle term of the GP is doubled, then the new numbers are in AP.

⇒ a, 2ar, ar² are in AP.

⇒ 4ar = a + ar²

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± √3

Since the series is an increasing GP, r = 2 + √3.

Example 8: Find the 9th and nth terms of a G.P 4, 16, 64 …

Solution:

Here, a = 4

r =  16/4 = 4

a₉ = ar^n-1 = 4 × 4^9-1 = 4 × 4⁸ = 4⁹

Also a_n = ar^n-1 = 4(4)^n-1 = 4ⁿ.

Example 9: Find the sum of the first 5 terms of the G.P 2, 8, 32, …

Solution:

Here, a = 2

r = 8/2= 4

n = 5

Sn  = a(rⁿ – 1)/(r – 1) 

= 2(4⁵ – 1)/(4 – 1)

= 2(4⁵ – 1)/3

= 682

Hence, the required sum is 682. 

Example 10: The geometric mean of two numbers is 6, and their arithmetic mean is 6.5. The numbers are

Solution:

Let a and b be the numbers.

GM = 6

√(ab) = 6

ab = 36

b = 36/a …(i)

AM = 6.5

(a + b)/2 = 6.5

(a + b) = 13 ..(ii)

Substitute b in (ii)

(a + 36/a) = 13

a² – 13a + 36 = 0

(a – 9)(a – 4) = 0

So a = 9 or a = 4

Hence, the numbers are 9 and 4.

Geometric Progression and Its Properties