if the 3rd and6th terms of a G.P. are 12 and 96, then find the number of terms in the progression, which are less than 2000
Answers
Answer:
Step-by-step explanation:
GP: a, ar, ar^2 , ar^3……….ar^n for n terms in a series.
Given that: ar^2=12 ……..(i)
ar^5=96……………(ii)
Dividing (ii) by (i) we get,
r^3=8
taking cuberoot of 8 gives us, r=2.
putting this value of r in (i) we get,
a*(2)^2=12
a=12/4
a=3.
Now lets suppose nth term in this series is less than 2000, which means that
a*r^(n-1) <= 2000
3*2^(n-1)<=2000
2^(n-1)<=2000/3
taking log (base 10) on both sides, we get
(n-1)log 2 <= log(2000/3)
on simplifying log we get
(n-1)*0.3010 <= 2.8239
on further solving,
(n-1) <= 9.38
n<=10.38
Thus the value of n should be less than 10.38 which implies that the 10th term in this series is less than 2000, while above 10 are greater than 2000.
Hence, the number of terms less than 2000 in this series are 10.