Question

For each geometric progression find the common ratio ‘r’, and then find an(i) 3, 3/2, 3/4, 3/8, ......... (ii) 2, −6, 18, −54(iii) −1, −3, −9, −18 .... (iv) 5, 2, 4/5, 8/25, .........

Answer 1

$a_1,a_2,a_3,......a_n$ are in Geometric progression only if $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}......=\frac{a_n}{a_{n-1}}=r$
and nth term in GP , $a_n=a_1r^{n-1}$

(i) 3, 3/2 , 3/4 , 3/8 ....
(3/2)/3 = (3/4)/(3/2) = (3/8)/(3/4) = 1/2
hence, common ratio , r = 1/2
nth term, $a_n=3(1/2)^{n-1}$

(ii) 2, -6, 18, -54 ....
-6/2 = 18/-6 = -54/18 = -3
hence, common ratio, r = -3
now, nth term ,$a_n=2(-3)^{n-1}$

(iii) -1, -3, -9, -18, ......
-3/-1 = -9/-3 ≠ -18/-9
hence, it is not in geometric progression.

(iv) 5, 2, 4/5 , 8/25 .....
2/5 = (4/5)/2 = (8/25)/(4/5) = 2/5
hence, common ratio, r = (2/5)
now, nth term , $a_n=5(2/5)^{n-1}$

Answer 2

Hi ,

******************************************

Let a , ar , ar² , ar³ ,... is a G.P

where a , r are first term and common

ratio

r = a2/a1 = an/an-1

nth term = an = ar^n - 1

*******************************************

Now ,

i ) 3 , 3/2 , 3/4 , 3/8 , ... is a G.P

a = a1 = 3 , a2 = 3/2 , a3 = 3/4

r = a2/a1 = ( 3/2 )/3 = ( 3/2 )× 3 = 9/2

an = ar^n-1

an = 3 × ( 9/2 )^n-1
______________________

ii ) 2 , -6 , 18, -54 ,....is given G.p

a = a1 = 2 , a2 = -6 ,

r = a2/a1 = ( -6 )/2 = -3

an = ar^n-1

an = 2 × ( -3 )^n-1
__________________________

iii ) -1 , -3 , -9 , -18 ,.... is given G.P

a = a1 = -1 , a2 = -3 ,

r = a2/a1 = ( -1 )/( -3 ) = 1/3

an = ar^n-1

an = ( -1 )× (1/3 )^n-1

________________________

iv ) 5 , 2 , 4/5 , 8/25 , ... is given G.P



a = a1 = 5 , a2 = 2 ,

r = a2/a1 = 2/5

an = ar^n-1

an = 5 × ( 2/5 )^n-1
________________________

I hope this helps you.

: )