80.5k views
1 vote
15. Find the sum of the first six terms of the geometric sequence for which a2 = 0.7 and a3 = 0.49.

16. Rewrite using radicals.


17. Rewrite using rational exponents.

User Felixwcf
by
8.3k points

1 Answer

5 votes

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.

User Douglasr
by
8.3k points

No related questions found