Question

Two terms in a geometric sequence are a3=729 and a4=243.

Answer 1

The first term of the given sequence (a) = 6561

Step-by-step explanation:

Let the first term = a and common difference = d

Given,

$a_{3}$ = 729 and $a_{4}$ = 243

To find, the first term of the given sequence (a) = ?

We know that,

The nth term of a G.P.

$a_{n} =ar^{n-1}$

The 3rd term of a G.P.

$a_{3} =ar^{3-1}$

⇒ $ar^{2}$ = 729          ..............(1)

The 4th term of a G.P.

$a_{4} =ar^{4-1}$

⇒ $ar^{3}$ = 243          ..............(2)

Dividing equation (2) by (1), we get

$\dfrac{ar^{3}}{ar^{2}}$ = $\dfrac{243}{729}$

⇒ $r=\dfrac{1}{3}$

Put $r=\dfrac{1}{3}$ in equation (1), we get

$a(\dfrac{1}{3})^{2}$ = 729    

⇒ $a(\dfrac{1}{9})$ = 729    

⇒ a = 9 × 729 = 6561

∴ The first term of the given sequence (a) = 6561