Question

How do you find the 7th term of the geometric sequence with the given terms a4 = 54, a5 = 162?

Answer 1

Solution:

Let a, r are first term and common ratio of an G.P

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

It is given that,

$a_{4}=54 \: a_{5}=162$

=> ar³ = 54 ---(1)

and

ar⁴ = 162 ----(2)

Do (2) ÷ (1) , we get

=> [(ar⁴)/(ar³)] = 162/54

=> r = 3

Put r = 3 in equation (1) , we get

=> a×3³ = 54

=> a × 27 = 54

=> a = 54/27

=> a = 2

Now ,

$a_{7} = ar^{6}$

= $2×3^{6}$

= $ 1458$

Therefore,

$a_{7} = 1458$

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