Question 935611: Please solve the 2nd term of an exponantial sequence which is 9 while the 4th term is 81. Find the common ratio,the first term and the sum of the first five terms of the sequence.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula for the exponential sequence is:
An = A0 * B^n
An is the nth term in the sequence.
A0 is the first term in the sequence
B is the base that is being raised to the exponent.
B is the common ratio.
n is the number of the term in the sequence.
n is also the exponent that B is being raised to.
for the second term in the sequence, you get:
A2 = A0 * B^2
since A2 is equal to 9, the equation becomes:
9 = A0 * B^2
for the fourth term in the sequence, you get:
A4 = A0 * B^4
since A4 is equal to 81, the equation becomes:
81 = A0 * B^4
your 2 equations are:
9 = A0 * B^2
81 = A0 * B^4
solve each of these equations for A0 and you get:
A0 = 9 / B^2
A0 = 81 / B^4
by substituting 9 / B^2 for A0 from the first equation, your second equation of A0 = 81 / B^4 becomes:
9 / B^2 = 81 / B^4
multiply both sides of this equation by B^4 and divide both sides of this equation by 9 and you get:
B^4 / B^2 = 81 / 9
simplify to get:
B^2 = 9
take the square root of both sides of this equation and you get:
B = plus or minus 3.
B can't be negative, so B has to be equal to 3.
that's your solution.
the common ratio, which is B, is equal to 3.
you can also solve for A1 by replacing B with 3.
from the above equations for A0, you have:
A0 = 9 / B^2
A0 = 81 / B^4
replace B with 3 and those equations becomes:
A0 = 9 / 3^2 = 9 / 9 = 1
A0 = 81 / 3^4 = 81 / 81 = 1
A1 is equal to 1.
B is equal to 3.
An is equal to 1 * 3^n
when n is equal to 0, An is equal to 1 * 3^0 which is equal to 1
when n is equal to 1, An is equal to 1 * 3^1 which is equal to 3
when n is equal to 2, An is equal to 1 * 3^2 which is equal to 9
when n is equal to 3, An is equal to 1 * 3^3 which is equal to 27
when n is equal to 4, An is equal to 1 * 3^4 which is equal to 81
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