Find the sum of the first six terms of the geometric sequence for which a2 = 0.7 and a3 = 0.49.
Let the common ratio of the geometric sequence be r. Then we have:
a2 = ar = 0.7
a3 = ar^2 = 0.49
Dividing the second equation by the first, we get:
r = a3/a2 = 0.49/0.7 = 7/10
So the first six terms of the sequence are:
a1 = a2/r = 0.7/(7/10) = 1
a2 = 0.7
a3 = 0.49
a4 = ar^2 = 0.343
a5 = ar^3 = 0.2401
a6 = ar^4 = 0.16807
The sum of the first six terms is:
S6 = a1 + a2 + a3 + a4 + a5 + a6
S6 = 1 + 0.7 + 0.49 + 0.343 + 0.2401 + 0.16807
S6 = 2.94017
Therefore, the sum of the first six terms of the geometric sequence is approximately 2.94017.
Rewrite using radicals.
The expression (x^3/4)^2 can be rewritten as:
(x^3/4)^2 = x^3/4 * x^3/4
(x^3/4)^2 = (x^3)^2 / 4^2
(x^3/4)^2 = x^6 / 16
So (x^3/4)^2 is equivalent to (x^6/16) using radicals.
Rewrite using rational exponents.
The expression √(a^3b^2) can be rewritten using rational exponents as:
√(a^3b^2) = (a^3b^2)^(1/2)
√(a^3b^2) = a^(3/2)b^(1)
So √(a^3b^2) is equivalent to a^(3/2)b^(1) using rational exponents.