SOLUTION: The second and fifth terms are in a geometric progression are 12 and 40.5 respectively. Find the first term and the common ratio. Hence, write down an expression for the nth term.

Algebra.Com
Question 1202406: The second and fifth terms are in a geometric progression are 12 and 40.5 respectively. Find the first term and the common ratio. Hence, write down an expression for the nth term.
Found 3 solutions by greenestamps, ikleyn, mananth:
Answer by greenestamps(13216)   (Show Source): You can put this solution on YOUR website!


Use the standard convention of a for the first term and r for the common ratio.

First term:
Second term:
Fifth term:

The ratio of the fifth and second terms is the common ratio, cubed. Use that to determine the common ratio.




The first term is the second term, divided by the common ratio.



ANSWERS:
1st term: 8
formula for n-th term (ar^(n-1):


Answer by ikleyn(52920)   (Show Source): You can put this solution on YOUR website!
.
The second and fifth terms in a geometric progression are 12 and 40.5 respectively.
Find the first term and the common ratio. Hence, write down an expression for the nth term.
~~~~~~~~~~~~~~~~~~~~~~

Write the terms in standard form

     = a*r,   = ,


Take the ratio

     =  =  =  = 3.375.


It implies  r =  = 1.5.


Thus the common ratio is  r= 1.5.


Now the first term is  a =  =  = 8.


The expression for the n-th term is   =  = .

Solved.     //     All questions are answered.

---------------------

On geometric progressions,  see introductory lessons
    - Geometric progressions
    - The proofs of the formulas for geometric progressions
    - Problems on geometric progressions
    - Word problems on geometric progressions
in this site.

Learn the subject from there.



Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
The second and fifth terms are in a geometric progression are 12 and 40.5 respectively. Find the first term and the common ratio. Hence, write down an expression for the nth term.
nth term of a GP is an=ar^(n-1)
2nd term
a2 = a*r^(2-1)
12= a*r
5th term
40.5= a*r^(5-1)
40.5 = ar^4
40.5= ar*r^3
40.5 = 12*r^3 (ar=12)
40.5/12 = r^3
3.375=r^3
3 15/40=r^3
135/40 =r^3
27/8 =r^3
cube root
r= 3/2

ar= 12
r = 3/2
a*3/2=12
a=8
an = a*r^(n-1)
an = 8*3/2 ^(n-1)


RELATED QUESTIONS

The first term of an arithmetic progression is 12 and the sum of the first 16 terms is... (answered by greenestamps)
A geometric progression and an arithmetic progression have the same first term. The... (answered by htmentor)
The second term and fifth term of a Geometric Progression (GP) are 1 and 1÷8... (answered by josgarithmetic)
If the third term and fifth term of a geometric progression are 225 and 5625... (answered by mananth)
if the first, fifth and tenth terms of an arithmetic sequence are in geometric... (answered by KMST)
The sixth and 13th terms of a geometric progression are 5/2 and 320 respectively.Find the (answered by greenestamps)
The sum of the first 100 terms of an arithmetic progression is 10000; the first, second... (answered by mananth)
IF 3(5/9) AND 40(1/2) ARE THE FIRST AND THE LAST TERMS RESPECTIVELY, OF A GEOMETRIC... (answered by KMST)
find the sum of the first 5 terms of the geometric progression for which the first and... (answered by richwmiller)