Question

in geometric progression the 2nd term is 3 and the 12th term is 729 then find the value of 10th term​

Answer 1

Answer:

let first term be a and common factor be x.

according to the question,

$a {x}^{2} = 3$

$a {x}^{12} = 729$

$\frac{(a {x}^{12}) }{(a {x}^{2}) } = \frac{729}{3}$

${x}^{10} = 243$

$x = \sqrt[10]{243} = \sqrt{3}$

so, the common factor is

$\sqrt{3}$

so, tenth term = 12th term/(√3)²

$\frac{729}{ { \sqrt{3} }^{2} }$

$\frac{729}{3}$

$243$

so, tenth term is 243.

Answer 2

Answer:

given in G.P, a2 = 3 and a12 = 729

to find a10

first we have to find first term a and common ratio r

a(n) = arⁿ-¹

=> a₂ = ar

=> a₁₂ = ar¹¹

now a₁₂/a₂ = ar¹¹/ar = 729/3

=> r¹⁰= 243

r = (243)¹/¹⁰

=> [(3)⁵]¹/¹⁰

=> 3¹/²

=> √(3)

ar = 3

a√(3) = 3

a = 3 /√3 ( by rationalising the denominator)

a = √3

therefore a = √3 and r = √3

10th term = ar⁹

=>√(3)(√3)⁹

=> √3 ( 81)√3

=> 243