Geometric Progression (G.P.)

Geometric Progression (G.P.)

A series in which each term is derived from the preceding by multiplying it by a constant factor is called a Geometric Progression or G.P. The constant factor is called the common ratio (CR) and is formed by dividing any term by the term which precedes it.

e.g.: 1, 2, 4, 8, 16.... or 3, 9, 27, 81, 243 ... etc.

The general form of a G.P. with n terms is a, ar, ar², ... a^{n-1 .} Thus, if a = the first term, r = the common ratio,

T_n = the n^th term,

Then T_n = a^{n -1}

E13.The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio of the G.P 

Sol. Let r be the common ratio of the G.P. It is given that the first term a = 1.

Now,a₃+a₅=90

ar² + ar⁴= 90

=r²+r⁴=90

r⁴ + r² —90 =0                                                                                      

r⁴ + 10r² — 9r² — 90 = 0.

     (r²+10)(r²-9)=0

      r² -9=0

     r=+ 3.

E14.The 3rd term of a G.P. is 36 and its 6th term is 288. What will be its 8th term?

Sol. Third term = ar² = 36                   ....(1)

Sixth term = ar⁵ = 288                         ....(2)

(2) divided by (1)=r³= 288/36 

=r³ =8=r=2.

Also eighth term = ar⁷ = ar⁵ x r² = 288 x 4 =1152.

Some Important Results

1.     If each term of a G.P. is multiplied or divided by a fixed non­zero constant, then the resulting sequence is also a G.P. e.g. 1, 3, 9, 27 .... is a G.P. If we multiply each term by 4, then the new sequence will be 4, 12, 36, 108 .... so the new sequence is also a G.P. with same common ratio of 3.

2.     If a_l, a₂, a₃ .... and b_l, b₂, b₃, .... are two geometric progressions, then the sequence a_l b_l, a₂ b₂, a₃, b₃ is also a G.P. e.g 4, 8, 16, 32 .... and 3, 9, 27, 81 .... are in G. P. If we multiply the corresponding terms of both G. Ps, then we will get new sequence 12, 72, 432, 2592 .... this is also a G. P with common ratio of 6. So in this process the new common ratio will be always the product of the two common ratios.

3.       The geometric mean G of two non-zero numbers a and b is given by √ab. It is to be noted that a, G and b are in G.P.

Important

If a_l, a₂....., a_n are n non-zero numbers, then their  mean is given by G = (a₁a₂ ...a _n)^1/n .

E16. Find the geometric mean between 2 and 32.

Sol. G. M = √ab

So G.M= √ 2x 32= √64  =8.

E17. Find the geometric mean of 1, 3 , 9 ,………, 1/3ⁿ-1.

Selection of terms in a G.P.

Sometimes it is required to select a finite number of terms in G.P.

It is always convenient if we select the terms in the following manner.

If the product of the numbers is not given, then the numbers are taken as a, ar, ar², ar³, ....

E18.The sum of three numbers in G.P. is 35 and their product is 1000. Find the numbers.

Sol. Let the numbers be a/r, a, ar.

Then, a/r • a • ar = 1000.

a³=1000

a = 10.

Also, a/r + a + ar = 35

= (10/r+10+10r) =35 [-.-a=10]

= 10(1+r+r²)=35r

= 2r² - 5r  + 2=0

(2r-1)(r-2)=0              

r = 1/2 or r = 2.

:. The required numbers are 20, 10, 5 or 5, 10, 20.

E19. Find a G.P of positive real numbers such that the product of its four terms is 16 and the common ratio is 2.

Sol. Let the four terms be a/r^{3, a/r} ,  ar , ar³.

Thena/r³   x a/r x ar x ar³ = 16

a⁴=16

a² = 4

a = ± 2.

Note that here the common ratio = r² = 2  

.•. r=±√2

The possible values of the 4 numbers are  1/√2, √2, 2√2, 4√2

5.        If a_l, a₂, a₃, ... is a G.P., then log a_l, log a₂, log a₃, ... is an A. P.

In this case the converse also holds good e.g.2, 4, 8, 16, ... are in G.P, then log 2, log 4, log 8, ... will be always in A. P.

Since log 2, log 4, log 8, ... can be written as log 2, log 2², log 2³, ... or log2, 2log2, 3log 2,....So it is an A.P. with a common difference of log 2.

6. The p numbers G_I, G₂, G₃, G₄, ..., G_P are said to be geometric means between a and b if a, G₁, G₂, ...,G_P, b is a G.P. Here a is the first term and b is the (p + 2)^th term of the G. P. If r is the common ratio of this G. P, then

E20. Find 3 geometric means between 1 and 16.

Sol. Let the geometric means be G₁, G₂ , G₃.

So 1, G₁, G₂, G₃, 16 will be in a G. P. Then 16 will be 5^th term of the G. P.

 Let the common ratio be r.

T₅ = 1.   r^5-1    ⟹             r⁴ = 16  ⟹      r=+ 2.

Then the possible means are 2, 4, 8 or -2, 4, -8.

Important

Ø In a finite G.P, the product of the terms equidistant from the beginning and the end is always same and is equal to the product of the first and the last term.

Ø If each term of a G.P. be raised to the same power, the resulting sequence also forms a G.P.

Ø The reciprocals of the terms of a given G.P. form a G.P.

The sum of n terms of a G.P.

The sum of n-terms of a G.P. with first term a and common ratio r is

       

 

E21. Find the sum to 20 terms of the G.P. 128, -96, 72, - 54,

Sol. Here a = 128, ar = -96, therefore, r = -3/4. Note that r < 1..

Solve yourself

Ans. 4²⁰ – 3²⁰ / 2³¹ . 7.