# 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 definition and properties of common ratio.

## Geometric Progression Definition

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

The properties of a geometric progression are:

• If a constant non zero quantity is multiplied or divided to each term, the resulting sequence is also a geometric progression.
• The reciprocal of all the terms in geometric progression also form 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:

 Representation of GP a, ar, ar², ar³, ……… arn-1, arn a = first term r = common ratio n = the last term in GP and,Sn = sum to n terms of GP The common ratio of GP $r=\frac{a_{n}}{a_{n-1}}$ The nth term of GP an = arn-1 Sum to n terms of a GP Sn = a (1−rn)/(1−r ) or Sn = a(rn -1)/(r-1)

## 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 $a_{1}=8\\ a_{2}=16\\ a_{3}=32\\ a_{4}=64\\ a_{5}=128\\ r=\frac{a_{2}}{a_{1}}=\frac{16}{8}=2\\ r=\frac{a_{4}}{a_{3}}=\frac{64}{32}=2\\$

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

Solution:

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

We know that nth 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 pth, qth and rth terms of a G.P., then what is (c / b)p (b / a)r (a / c)q equal to?

Solution:

a = ARp−1, b = ARq−1, c = ARr−1

(c / b)p * (b / a)r * (a / c)q = [(ARr−1 / ARq−1) ]p * [(ARq−1 / ARp−1)]r * [(ARp−1/ ARr−1)]q

= R(r−q)p+(q−p)r+(p−r)q

= R0

= 1

Example 3: The 6th term of a G.P. is 32 and its 8th 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 series?

Solution:

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

Sum of n terms of series is Sn= 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 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 $a,ar,ar^{2}$ be in GP.

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

$\Rightarrow \text \ a,\ 2ar,\ ar^{2} \text \ are \ in \ AP\\ \Rightarrow 4ar=a+ar^{2}\\ \Rightarrow r^{2}-4r+1=0\\ \Rightarrow r=2\pm \sqrt{3}\\$

Since the series is an increasing GP, $r=2+\sqrt{3}$

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

Solution:

Here a = 4

r =  16/4 = 4

a9 = arn-1 = 4×49-1 = 4×48 = 49

Also an = arn-1 = 4(4)n-1 = 4n.

Example 9: Find sum of  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.