Find the G. P. in which 4th term is 3 and the 7th term is 8/9
Answers
Answer:
4th term is 3 ---> ar^3 = 3
7th term is 8/9 --> ar^6 = 8/9
divide them:
r^3 = 8/27
r = 2/3
in ar^3 = 3
a(8/27) = 3
a = 81/8
GP is 81/8 , 27/4 , 9/2 , 3 , .....
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Therefore the series of Geometric Progression is 81/8, 27/4, 9/2, 3,...
Given:
The 4th term in the series = 3
The 7th term in the series = 8/9
To Find:
The series of Geometric progression.
Solution:
The given question can be solved as shown below.
Let a = First term in the series
r = common difference
Given that,
The 4th term in the series = 3
The 7th term in the series = 8/9
The nth term series in the GP series = arⁿ⁻¹
⇒ The 4th term in the series = arⁿ⁻¹ = ar³ = 3 __(i.)
⇒ The 7th term in the series = arⁿ⁻¹ = ar⁶ = 8/9__(ii.)
Dividing equation (ii.) by (i.)
⇒ ar⁶/ar³ = (8/9) / 3
⇒ r³ = 8/27 ⇒ r = 2/3
From equation-(i.),
⇒ ar³ = 3
⇒ a = 3/(2/3)³ = 81/8
So the first term of the series = a = 81/8
The 2nd term in the series = ar = (81/8) × (2/3) = 27/4
The 3rd term in the series = ar² = (81/8) × (2/3)² = 9/2
GP series = a, ar, ar², ar³......
GP series = 81/8, 27/4, 9/2, 3,....
Therefore the series of Geometric Progression is 81/8, 27/4, 9/2, 3,...
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