Find the $$12^{th}$$ term of the G.P. $$\dfrac{1}{a^3x^3} , ax, a^5x^5, …$$

Asked by Sakshi | 1 year ago |  140

##### Solution :-

We know that,

t1 = a =$$\dfrac{ 1}{a^3x^3}$$, r = $$\dfrac{ ax}{a^3x^3}$$ = ax (a3x3) = a4x4

By using the formula,

Tn = arn-1

T12 = $$\dfrac{ 1}{a^3x^3 (a^4x^4)^{12-1}}$$

$$\dfrac{ 1}{a^3x^3 (a^4x^4)^{11}}$$

= (ax)41

Answered by Sakshi | 1 year ago

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.