Find the number of terms in G.Pgiven thatits first&last term are 5 1/3k&243k/256 respectively&that its common ratiois3/4
first term = a = 5 1/3k or (16/3)k
r = 3/4
last term = ar^(n-1) = 243k/256 = (243/256)k
(16/3)k (3/4)^(n-1) = (243/256)k
(3/4)^(n-1) = (243/256)(3/16) = 729/4096 = 3^6 / 4^6
(3/4)^(n-1) = (3/4)^6
n-1 = 6
n = 7
you have 7 terms
Very nice 👌👌
Correct
please I don't understand how you got 6
Correct
Confirmed
To find the number of terms in a geometric progression (G.P.), you'll need to use the formula for the nth term of a G.P., which is:
an = a * r^(n-1)
Where:
an = nth term of the G.P.
a = first term of the G.P.
r = common ratio of the G.P.
n = number of terms in the G.P.
In this case, the first term (a) is 5 1/3k, and the last term is 243k/256. The common ratio (r) is 3/4.
Let's start by finding the values of a and r:
a = 5 1/3k
We can write 5 1/3 as an improper fraction: 16/3.
So, a = (16/3)k.
Now, let's find r:
r = 3/4
Next, we want to find the value of n. We know that the nth term (an) is equal to 243k/256. So, we can substitute these values into the formula:
(243k/256) = [(16/3)k] * [(3/4)^(n-1)]
To simplify this equation, we can cancel out the k terms:
(243/256) = (16/3) * [(3/4)^(n-1)]
Now, let's simplify the equation further:
(243/256) = (16/3) * [(3^(n-1))/(4^(n-1))]
To eliminate the fractions, we can multiply both sides of the equation by 256 * 3 * 4^(n-1) to get rid of the denominators:
(243/256) * (256 * 3 * 4^(n-1)) = (16/3) * [(3^(n-1))/(4^(n-1))] * (256 * 3 * 4^(n-1))
Simplifying further:
243 * 3 * 4^(n-1) = 16 * (3^(n-1)) * (256 * 3 * 4^(n-1))
Canceling out terms:
729 * 4^(n-1) = 768 * 3 * (4^(n-1))
Dividing both sides by 3 to simplify:
729 * 4^(n-1) / 3 = 768 * 4^(n-1)
Now, we can cancel out the 4^(n-1) terms:
729 / 3 = 768
243 = 768
This equation doesn't hold true, which means that there is no solution. Therefore, there is no whole number value for n that satisfies this equation. Hence, there are no terms in the given G.P.
Please check the values provided or the calculations to ensure accuracy.