**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, ar^{2}, ar^{3},...................

Here a = first term and r - t_{2}/t_{1}

**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 = ar^{2} = 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 = ar^{2} = √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 = ar^{2} = 1000(2/5)^{2} = 1000(4/25) = 160

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

n^{th} term of a geometric sequence :

t_{n} = ar ^{n -1}

**Question 1 :**

In a G.P. 729, 243, 81,… find t_{7} .

**Solution :**

t_{n} = ar ^{n -1}

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

t_{7} = (729) (1/3)^{7 -1}

= (729) (1/3)^{6}

t_{7} = 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)

x^{2} + 12^{2} + 2x(12) = x^{2} + 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 n^{th} term be "8192"

t_{n} = 8192

a = 4, r = 8/4 = 2

ar ^{n -1 } = 8192

4(2)^{n -1 } = 8192

2^{2}(2)^{n -1 } = 8192

2 ^{n+1} = 2^{13}

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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