Question

6. If the geometric sequences 162, 54, 18,g. and , , , 2\81 2\27 2\9 g have their nth term
equal, find the value of n.

Answer 1

GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)

a1 = a , r = a(n+1)/ an

General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........

SOLUTION :
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

Hence, the value of n = 5

HOPE THIS WILL HELP YOU….

Answer 2

Solution :

i) 162 , 54 , 18 , ... , are in G.P

first term = a = a1 = 162

Common ratio = r = a2/a1

r = 54/162 = 1/3

nth term = an = ar^n-1

an = 162 × ( 1/3 )^n-1 ----( 1 )

ii ) 2/81 , 2/27 , 2/9, ...,are in G.P

a = a1 = 2/81 ,

r = ( 2/27 )/( 2/81 ) = 81/27 = 3

nth term = ( 2/81 ) × 3^n-1 ----( 2 )

iii ) According to the problem given ,

( 1 ) = ( 2 )

162 × ( 1/3 )^n-1 = ( 2/81 ) × 3^n-1

=> 162 × ( 81/2 ) = 3^n-1 × 3^n-1

=> 81² = ( 3^n-1 )²

=> 81 = 3^n-1

=> 3⁴ = 3^n-1

4 = n - 1

[ Since , If a^m = a^n then m = n ]

n = 5

Therefore ,

n = 5

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