the 3rd and 6th term of a geometric progression are 48 and 14 2/9 respectively.
a1, a2, a3=48, a4, a5, a6=
You need this formula for the nth term:
an = a1·r(n-1)
[Some books and teachers use "t" for "term" instead of "a".
I'll use "a".]
the 3rd and 6th term of a geometric progression are 48 and 14 2/9 respectively.
a3 = a1·r(3-1)
48 = a1·r2
a6 = a1·r(6-1)
= a1·r5
Change to an improper fraction
= a1·r5
Clear of fractions by multipolying both sides by 9
128 = 9a1·r5
So you have this system of equations:
48 = a1·r2
128 = 9a1·r5
Solve the first for a1
= a1
Substitute in
128 = 9·r5
128 =
Divide on the right by subtracting exponents:
128 = 432r³
Divide both sides by 432
=
Reduce the fraction on the left and cancel on the right:
=
= r³
The cube root of 8 is 2 and the cube root of 27 is 3, so
taking cube roots of both sides:
= r
Substitute for r in
= a1
= a1
= a1
To divide by a fraction, invert it and multiply
= a1
4 goes into 48 12 times:
12
= a1
1
108 = a1
Write down the 1st 4 terms of the geometric progression.
1st term = ,
2nd term =
3rd term =
4th term =
the 3rd term checks, but to check the whole thing,
lets see if the 6th term is
5th term =
6th term =
So it checks.
Edwin