Question

3. If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, find the G.P

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 :
4th term = 54

7th term = 1458
t₄ = 54
tn = arⁿ-1
a r³ = 54 ----- (1)

t₇ = 1458
a r⁶ = 1458 ----- (2)

On dividing eq 2 by 1
(a r⁶)/(a r³) = 1458/54
r³ = 27
r³ = 3³
r = 3

Put r = 3 in the equation 1, we get
a r³ = 54
a (3)³ = 54
a(27) = 54
a = 54/27
a = 2

General form of G.P is a, a r , a r ²,.........

2 ,2(3),2(3)²,......

2,6,18,.......


Hence, the G.P is 2,6,18…

HOPE THIS WILL HELP YOU….

Answer 2

Solution :

Let a , r are the first term and

Common ratio of a G.P .

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nth term of G.P = an

an = ar^n-1

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It is given that

i ) a4 = 54

=> ar³ = 54 ----( 1 )

ii ) 7th term = 1458

=> ar^6 = 1458 ----( 2 )

Do ( 2 ) ÷ ( 1 ) , we get

[ ar^6/ar³ ] = 1458/54

=> r³ = 27

=> r³ = 3³

r = 3

Substitute r = 3 in ( 1 ) , we get

a × 3³ = 54

=> 27a = 54

=> a = 54/27

a = 2

Therefore ,

Required G.P is

2 , 6 , 18 , 54 ,....

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