Question

7.
If the
geometric progressions 162, 54, 18 .....and 2/81,2/27,2/9........have their nth term equal .Find the value of n​

Answer 1

Step-by-step explanation:

GIVEN : 1st sequence

162,54,18,.........

Here, a = 162

r = a(n+1)/ an

r = 54/162

r = 1/3

2nd sequence

2/81,2/27,2/9,..........

Here, a = 2/81

r = a(n+1)/ an

r = (2/27)/ (2/81)

r = 2/27 x 81/2

r = 3

tn = arⁿ-1

tn = 162 (1/3)ⁿ-1

tn = 162 (3⁻¹)ⁿ-1

= 162 (3⁻ⁿ⁺¹) ………… (1)

tn = arⁿ-1

tn = (2/81) x (3)ⁿ-1 ……… (2)

t n = t n

Since nth term of the given geometric sequence are equal

162 (3⁻ⁿ⁺¹) = (2/81) x (3)ⁿ-1

[ From eq 1 & 2]

[(162 x 81)/2] (3⁻ⁿ⁺¹) = (3)ⁿ-1

(81 x 81)(3⁻ⁿ⁺¹) = (3)ⁿ-1

(3⁴ x 3⁴)(3⁻ⁿ⁺¹) = (3)ⁿ-1

(3⁸)(3⁻ⁿ⁺¹) = (3)ⁿ-1

(3)⁻ⁿ ⁺ ⁹ = (3)ⁿ-1

- n + 9 = n - 1

- n - n = -1 - 9

-2 n = -10

n = 10/2

n = 5