Question

the 4th term of a geometric sequence is 2/3 and the 7th term is 81/16 find the geometric series
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Answer 1

Solution :-

We know that nth term of a geometric progression is arⁿ-¹

⇒ T₄ = ar³ =2/3 ...... ( i )

⇒ T₇ = ar⁶ = 16/81 ...... ( ii )

On dividing ( ii ) by ( i ) , we get

⇒ ar⁶ / ar³ = 16/81 ÷ 2/3

⇒ r³ = 16/81 ÷ 2/3

⇒ r³ = 16/81 × 3/2

⇒ r³ = 8/27

⇒ r = 2/3 .

Now

⇒ ar³ = 2/3

⇒ a = 2/3 ÷ 8/27

⇒ a = 2/3 * 27/8 = 9/4 .

⇒ ar = 9/4(2/3) = 3/2

⇒ ar² = 9/4(4/9) = 1

⇒ ar³ = 2/3

G.P :- 9/4 , 3/2 , 1 ,2/3 ....

Answer 2

Step-by-step explanation:

Geometric series = a,ar,ar2,ar3......ar(n−1)

The fourth term of a geometric series = ar3=23

The Seventh term of a geometric series = ar6=1681

=> ar6ar3=1681×32

=> r3=827

= > ar3=23

=> a(827)=23

=> a 23×278=94

The first term of the geometric series is 9/4