If the third and ninth temr of a geometric series with a positive common ratio are -3 and -192 repectively, determine the value of

Question

If the third and ninth temr of a geometric series with a positive common ratio are -3 and -192 repectively, determine the value of

the first term a
Let a_1 be denoted by 'a' (first terms) then,
a_3=r^2 a
And,
a_9=r^8 a
Thus,
r^8 a=-192
r^2 a=-3
Divide two equations,
r^6=64=2^6
Thus,
r=2

The third term is,
-3=2^2 a
Thus,
a=-3/4

You used the ratio of t_n/t_m = r^n-m, where n>m. In this case n=9 and m=3.
It looks like you're getting a good handle on geometric sequences. Well done!

Answer 1

Thank you! I'm glad you found my explanation helpful.

To determine the value of the first term 'a', we can use the information given about the third and ninth terms of the geometric series. We know that the ratio of any two consecutive terms in a geometric series is constant, and we can use this information to find the value of 'a'.

Let's denote the first term as 'a', and the common ratio as 'r'.

We are given that the third term is -3, which means a₃ = -3. Using the formula for the nth term of a geometric series, we can express this as a₃ = a * r² = -3.

Similarly, we are given that the ninth term is -192, which means a₉ = -192. Using the same formula, we can express this as a₉ = a * r⁸ = -192.

Now, we can solve these two equations simultaneously to find the values of 'a' and 'r'.

Dividing the second equation by the first equation, we get:

(r⁸ * a) / (r² * a) = -192 / -3

Simplifying further, we have:

r⁶ = 64

Since 64 = 2⁶, we can deduce that r = 2.

Now that we know the value of 'r', we can substitute it back into either equation to find the value of 'a'. Let's use the equation a₃ = a * r² = -3:

a * (2²) = -3

4a = -3

Dividing both sides by 4, we find:

a = -3/4

Therefore, the value of the first term 'a' is -3/4.