HOW TO FIND THE FIRST THREE TERMS OF A GEOMETRIC SEQUENCE

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)  =  2√2

Hence the first three terms are √2, 2, 2√2

(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 ?

Question 2 :

In a G.P. 729, 243, 81,… find t₇ .

Solution :

t_n  =  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 3 :

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 4 :

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

(i) 4, 8, 16,…,8192 ?

Solution :

Let n^th term be "8192"

t_n  =  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 12^th term of the above geometric sequence is 8192.

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

Solution :

Let n^th term be "1/2187"

t_n  =  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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