HOW TO FIND THE FIRST THREE TERMS OF A GEOMETRIC SEQUENCE

How to Find the First Three Terms of a Geometric Sequence ?

A Geometric Progression is a sequence in which each term is obtained by multiplying a fixed non-zero number to the preceding term except the first term. The fixed number is called common ratio. The common ratio is usually denoted by r.

General form of geometric progression :

a, ar, ar2, ar3,...................

Here a = first term and r - t2/t1

How to Find the First Three Terms of a Geometric Sequence - Questions

Question 1 :

Write the first three terms of the G.P. whose first term and the common ratio are given below.

(i) a = 6, r = 3

Solution :

First term (a)  =  6

Second term  =  ar  =  6(3)  =  18

Third term  =  ar2  =  6(3)2  =  54

Hence the first three terms are 6, 18, 54.

(ii) a = 2, r = √2

Solution :

First term (a)  =  2

Second term  =  ar  =  2(2)  =  2

Third term  =  ar2  =  2(2)2  =  2(2)  =  22

Hence the first three terms are 2, 2, 22

(iii)  a = 1000, r = 2/5

Solution :

First term (a)  =  1000

Second term  =  ar  =  1000(2/5)  =  400

Third term  =  ar2  =  1000(2/5)2  =  1000(4/25)  =  160

Hence the first three terms are 1000, 400, 160.

How to Find the Indicated Term of a Geometric Sequence ?

nth term of a geometric sequence :

tn  =  ar n -1

Question 1 :

In a G.P. 729, 243, 81,… find t7 .

Solution :

tn  =  ar n -1

a = 729, r = 243/729  =   1/3 and n = 7

t7  =  (729) (1/3)7 -1

 =  (729) (1/3)6

t7   =  1

Question 2 :

Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression.

Solution :

b = √ac

(x + 12)  =  √(x + 6) (x + 15)

Taking squares on both sides,

(x + 12)2  =  (x + 6) (x + 15)

x2 + 122 + 2x(12)  =  x2 + 15x + 6x + 90

144 + 24x  =  21x + 90

24x - 21x  =  90 - 144

3x  =  -54

x  =  -18

Hence the value of x is -18.

Question 3 :

Find the number of terms in the following G.P.

(i) 4, 8, 16,…,8192 ?

Solution :

Let nth term be "8192"

tn  =  8192

a = 4, r = 8/4  =  2

 ar n -1  =  8192

4(2)n -1  =  8192

22(2)n -1  =  8192

2 n+1  =  213

n + 1  =  13

n = 12

Hence the 12th term of the above geometric sequence is 8192.

(ii) 1/3, 1/9, 1/27,................1/2187

Solution :

Let nth term be "1/2187"

tn  =  1/2187

a = 1/3, r = (1/9)/(1/3)  =  1/3

 ar n -1  =  1/2187

(1/3)(1/3)n -1  =  1/2187

(1/3)1 + n-1  =  1/2187

(1/3)n  =  (1/3)7

n  =  7

Hence the 7th term of the above sequence is 1/2187.

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