If the 4th and 9th terms of a GP are 54 & 13122 respectively, find the GP.
Answers
If the 4th and 9th terms of a GP are 54 & 13122 respectively, find the GP.
Let first term be a and common difference be r.
Given a4 = 54, a9 = 13122.
=) a× r^(4-1) = 54
=) ar³ = 54 - - 1)
a9 = 13122
=) a × r^(9-1) = 13122
=) ar^8 = 13122 - - 2)
Divide eq1 by eq2)
=) ar^8/ar^3 = 13122/54
=) r^5 = 243
=) r^5 = 3^5
=) r = 3.
Putting r in eq1)
=) a(3)³ = 54
=) a = 2.
Hence required GP is 2, 6, 18,54....
Heya dear !
The general formula for the nth term of a geometric progression is:
a(n) = ar^(n-1)
a : first term
r : common ratio
So, the terms of a geometric progression are:
a, ar, ar², ar^3, ar^4, etc.
Fourth term:
ar^3 = 54
Ninth term:
ar^8 = 13122
Dividing the 9th term by the 4th term:
ar^8 / ar^3 = 13122 / 54
r^5 = 243
r = 243^(1/5)
r = 3
Now we just need to figure the first term:
ar^3 = 54
Plug in r=3:
a(3^3) = 54
27a = 54
a = 54/27
a = 2
Answer:
a(n) = 2*3^(n-1)
2, 6, 18, 54, 162, 486, etc.
Hope this helps (:
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