Episode 8 — Aptitude and Reasoning / 8.17 — Geometric Progression

8.17.c Geometric Progression - Solved Examples

Example 1: Finding the Common Ratio

Problem: Find the common ratio of the GP: 5, 15, 45, 135, ...

Solution:

r = a(2)/a(1) = 15/5 = 3

Verification:
45/15 = 3
135/45 = 3

Common ratio r = 3

Example 2: Finding the nth Term

Problem: Find the 9th term of the GP: 2, 6, 18, 54, ...

Solution:

a = 2, r = 6/2 = 3, n = 9

a(n) = a * r^(n-1)
a(9) = 2 * 3^(9-1)
     = 2 * 3^8
     = 2 * 6561
     = 13122

The 9th term is 13,122.

Example 3: Finding Position of a Term

Problem: Which term of the GP 3, 12, 48, 192, ... is 12,288?

Solution:

a = 3, r = 12/3 = 4, a(n) = 12288

a(n) = a * r^(n-1)
12288 = 3 * 4^(n-1)
4096 = 4^(n-1)
4^6 = 4^(n-1)
n - 1 = 6
n = 7

12,288 is the 7th term.

Example 4: Sum of First n Terms (r > 1)

Problem: Find the sum of the first 8 terms of the GP: 3, 6, 12, 24, ...

Solution:

a = 3, r = 2, n = 8

S(n) = a(r^n - 1)/(r - 1)
S(8) = 3(2^8 - 1)/(2 - 1)
     = 3(256 - 1)/1
     = 3 * 255
     = 765

Example 5: Sum of First n Terms (r < 1)

Problem: Find the sum of the first 6 terms of the GP: 81, 27, 9, 3, ...

Solution:

a = 81, r = 27/81 = 1/3, n = 6

S(n) = a(1 - r^n)/(1 - r)
S(6) = 81(1 - (1/3)^6)/(1 - 1/3)
     = 81(1 - 1/729)/(2/3)
     = 81(728/729) * (3/2)
     = 81 * 728/729 * 3/2
     = 728/9 * 3/2
     = 728 * 3 / (9 * 2)
     = 2184/18
     = 364/3
     = 121.33...

Alternatively:
S(6) = 81 + 27 + 9 + 3 + 1 + 1/3 = 121 + 1/3 = 364/3

Example 6: Sum to Infinity

Problem: Find the sum to infinity of the GP: 16, 8, 4, 2, 1, ...

Solution:

a = 16, r = 8/16 = 1/2

Since |r| = 1/2 < 1, sum to infinity exists.

S(infinity) = a/(1 - r)
            = 16/(1 - 1/2)
            = 16/(1/2)
            = 32

Example 7: Sum to Infinity with Negative Ratio

Problem: Find the sum: 27 - 9 + 3 - 1 + 1/3 - ...

Solution:

a = 27, r = -9/27 = -1/3

|r| = 1/3 < 1, so sum exists.

S(infinity) = a/(1 - r)
            = 27/(1 - (-1/3))
            = 27/(1 + 1/3)
            = 27/(4/3)
            = 27 * 3/4
            = 81/4
            = 20.25

Example 8: Recurring Decimal as Fraction

Problem: Express 0.454545... as a fraction.

Solution:

0.454545... = 45/100 + 45/10000 + 45/1000000 + ...

This is a GP with:
  a = 45/100, r = 1/100

S(infinity) = (45/100)/(1 - 1/100)
            = (45/100)/(99/100)
            = 45/99
            = 5/11

0.454545... = 5/11

Example 9: Three Numbers in GP

Problem: The sum of three numbers in GP is 21 and their product is 216. Find the numbers.

Solution:

Let the numbers be a/r, a, ar

Product = (a/r)(a)(ar) = a^3 = 216
a^3 = 216
a = 6

Sum = a/r + a + ar = 21
6/r + 6 + 6r = 21
6/r + 6r = 15

Multiply by r:
6 + 6r^2 = 15r
6r^2 - 15r + 6 = 0
2r^2 - 5r + 2 = 0
(2r - 1)(r - 2) = 0
r = 1/2 or r = 2

If r = 2: Numbers are 6/2, 6, 6*2 = 3, 6, 12
If r = 1/2: Numbers are 12, 6, 3

The numbers are 3, 6, 12.

Example 10: GP from Two Terms

Problem: The 3rd term of a GP is 24 and the 6th term is 192. Find the GP.

Solution:

a(3) = ar^2 = 24     ... (i)
a(6) = ar^5 = 192    ... (ii)

Divide (ii) by (i):
r^3 = 192/24 = 8
r = 2

From (i): a * 4 = 24 -> a = 6

GP: 6, 12, 24, 48, 96, 192, ...

Example 11: Inserting Geometric Means

Problem: Insert 4 geometric means between 3 and 96.

Solution:

a = 3, b = 96, n = 4

r = (b/a)^(1/(n+1)) = (96/3)^(1/5) = 32^(1/5) = 2

G(1) = 3 * 2 = 6
G(2) = 3 * 4 = 12
G(3) = 3 * 8 = 24
G(4) = 3 * 16 = 48

Complete GP: 3, 6, 12, 24, 48, 96

Verification: Each term is double the previous. Correct!

Example 12: Geometric Mean

Problem: Find the geometric mean of 4 and 25.

Solution:

GM = sqrt(a * b) = sqrt(4 * 25) = sqrt(100) = 10

The geometric mean is 10.

Verification: 4, 10, 25
10/4 = 2.5 and 25/10 = 2.5 (same ratio). Correct!

Example 13: AM-GM Inequality Application

Problem: If x + y = 10 and x, y > 0, find the maximum value of xy.

Solution:

By AM-GM inequality:
AM >= GM
(x + y)/2 >= sqrt(xy)
10/2 >= sqrt(xy)
5 >= sqrt(xy)
25 >= xy

Maximum value of xy = 25 (achieved when x = y = 5)

Example 14: Bouncing Ball Problem

Problem: A ball is dropped from a height of 80 meters. Each time it hits the ground, it bounces to 3/4 of the previous height. Find the total distance traveled before it comes to rest.

Solution:

h = 80 m, r = 3/4

Total distance = h + 2h*r + 2h*r^2 + 2h*r^3 + ...
               = h + 2hr(1 + r + r^2 + ...)
               = h + 2hr/(1 - r)
               = h + 2h * (3/4) / (1 - 3/4)
               = 80 + 2 * 80 * (3/4) / (1/4)
               = 80 + 2 * 80 * 3
               = 80 + 480
               = 560 meters

Alternative formula:
Total = h(1 + r)/(1 - r) = 80 * (1 + 3/4)/(1 - 3/4)
      = 80 * (7/4)/(1/4) = 80 * 7 = 560 meters

Example 15: Finding Sum from nth Term

Problem: If the nth term of a GP is a(n) = 5 * 3^(n-1), find the sum of the first 10 terms.

Solution:

a(n) = 5 * 3^(n-1)

So: a = 5, r = 3

S(10) = 5(3^10 - 1)/(3 - 1)
      = 5(59049 - 1)/2
      = 5 * 59048/2
      = 5 * 29524
      = 147620

Example 16: Sum of Infinite GP Given Sum and First Term

Problem: The sum of an infinite GP is 20 and the first term is 4. Find the common ratio.

Solution:

S(infinity) = a/(1 - r) = 20
4/(1 - r) = 20
1 - r = 4/20 = 1/5
r = 1 - 1/5 = 4/5

Common ratio r = 4/5

Verification: |4/5| < 1, so infinite sum exists.
S = 4/(1 - 4/5) = 4/(1/5) = 20. Correct!

Example 17: Product of Terms in GP

Problem: Find the product of the first 5 terms of the GP: 4, 12, 36, 108, 324.

Solution:

Method 1: Direct multiplication
4 * 12 * 36 * 108 * 324

Method 2: Using the formula
Product = (a1 * a5)^(n/2) = (4 * 324)^(5/2) = 1296^(5/2)

Hmm, let's use the middle term method (for odd n):
n = 5, middle term = a(3) = 36
Product = 36^5 = 36 * 36 * 36 * 36 * 36

36^2 = 1296
36^3 = 46656
36^4 = 1679616
36^5 = 60466176

Verification: 4 * 12 = 48, 48 * 36 = 1728, 1728 * 108 = 186624, 186624 * 324 = 60466176. Correct!

Example 18: GP with Negative Ratio

Problem: Find the 7th term and sum of first 7 terms of GP: 1, -2, 4, -8, ...

Solution:

a = 1, r = -2

a(7) = 1 * (-2)^6 = 1 * 64 = 64

S(7) = a(r^7 - 1)/(r - 1)
     = 1((-2)^7 - 1)/(-2 - 1)
     = (-128 - 1)/(-3)
     = -129/(-3)
     = 43

Example 19: Logs and GP

Problem: If a, b, c are in GP, prove that log a, log b, log c are in AP.

Solution:

Given: a, b, c are in GP
Therefore: b^2 = ac

Taking logarithm on both sides:
2 log b = log a + log c

This means: log b - log a = log c - log b

Since the difference between consecutive terms is equal,
log a, log b, log c are in AP.

Common difference of the AP = log b - log a = log(b/a) = log(r)

Example 20: Compound Amount as GP

Problem: Rs. 10,000 is invested at 10% compound interest per annum. Find the amount after 5 years.

Solution:

Principal = 10,000, Rate = 10% = 0.1

Amount after each year forms a GP:
  After year 1: 10,000 * 1.1 = 11,000
  After year 2: 10,000 * 1.1^2 = 12,100
  ...
  After year n: 10,000 * 1.1^n

This is a GP with a = 11,000 and r = 1.1

After 5 years:
Amount = 10,000 * (1.1)^5
       = 10,000 * 1.61051
       = Rs. 16,105.10

Example 21: Sum of Specific Terms

Problem: In a GP with first term 5 and common ratio 2, find the sum from the 4th term to the 10th term.

Solution:

a = 5, r = 2

Sum from 4th to 10th = S(10) - S(3)

S(10) = 5(2^10 - 1)/(2-1) = 5(1024 - 1) = 5 * 1023 = 5115
S(3) = 5(2^3 - 1)/(2-1) = 5(8 - 1) = 5 * 7 = 35

Sum from 4th to 10th = 5115 - 35 = 5080

Example 22: Ratio of GP Sums

Problem: In a GP, S(3) = 13 and S(6) = 364. Find the common ratio.

Solution:

S(3) = a(r^3 - 1)/(r - 1) = 13       ... (i)
S(6) = a(r^6 - 1)/(r - 1) = 364      ... (ii)

Divide (ii) by (i):
(r^6 - 1)/(r^3 - 1) = 364/13 = 28

Note: r^6 - 1 = (r^3 - 1)(r^3 + 1)

So: (r^3 + 1) = 28
    r^3 = 27
    r = 3

Verification from (i): a(27-1)/(3-1) = a*26/2 = 13a = 13 -> a = 1
Check S(6) = 1(729-1)/2 = 728/2 = 364. Correct!

Example 23: AGP Problem

Problem: Find the sum to infinity: 1 + 3/2 + 5/4 + 7/8 + 9/16 + ...

Solution:

This is an AGP where:
  AP part: 1, 3, 5, 7, 9, ... (a = 1, d = 2)
  GP part: 1, 1/2, 1/4, 1/8, 1/16, ... (r = 1/2)

Let S = 1 + 3/2 + 5/4 + 7/8 + 9/16 + ...

Multiply by r = 1/2:
S/2 = 1/2 + 3/4 + 5/8 + 7/16 + 9/32 + ...

Subtract:
S - S/2 = 1 + (3/2 - 1/2) + (5/4 - 3/4) + (7/8 - 5/8) + ...
S/2 = 1 + 2/2 + 2/4 + 2/8 + 2/16 + ...
S/2 = 1 + 1 + 1/2 + 1/4 + 1/8 + ...
S/2 = 1 + [1/(1 - 1/2)]
S/2 = 1 + 2 = 3
S = 6

Alternatively using formula: S = a/(1-r) + dr/(1-r)^2
= 1/(1/2) + 2*(1/2)/(1/2)^2 = 2 + 1/(1/4) = 2 + 4 = 6

Example 24: Mixed AP-GP Problem

Problem: If the AM between two numbers is 25 and their GM is 7, find the numbers.

Solution:

AM = (a + b)/2 = 25 -> a + b = 50
GM = sqrt(ab) = 7 -> ab = 49

a and b are roots of: x^2 - 50x + 49 = 0
(x - 1)(x - 49) = 0
x = 1 or x = 49

The numbers are 1 and 49.

Verification:
AM = (1 + 49)/2 = 25. Correct!
GM = sqrt(1 * 49) = 7. Correct!

Next: 8.17 - Practice MCQs