The first and last terms of a Ap are 6.7 and 17.1respectively .if there are 14 term in the sequence.find the common difference
17.1 = 6.7 + 13d
Why did the arithmetic progression go to therapy?
Because it couldn't find its common difference!
*ba dum tss*
Alright, let's solve this.
The formula for the nth term of an arithmetic progression is:
An = A1 + (n - 1)d
Where An is the nth term, A1 is the first term, n is the number of terms, and d is the common difference.
We know that A1 = 6.7 and An = 17.1, and n = 14.
So, plugging in the values into the formula:
17.1 = 6.7 + (14 - 1)d
17.1 = 6.7 + 13d
10.4 = 13d
Dividing both sides by 13, we get:
d ≈ 0.8
So, the common difference is approximately 0.8.
To find the common difference of an arithmetic progression (AP), we can use the formula:
Common difference (d) = (last term - first term) / (number of terms - 1)
Given that the first term is 6.7, the last term is 17.1, and there are 14 terms in the sequence, we can substitute these values into the formula:
d = (17.1 - 6.7) / (14 - 1)
Calculating this expression yields:
d = 10.4 / 13
Simplifying further:
d ≈ 0.8
Therefore, the common difference of the arithmetic progression (AP) is approximately 0.8.
To find the common difference in an arithmetic progression (AP), we can use the formula:
nth term = first term + (n - 1) * common difference
In this case, we are given the first term (6.7) and the last term (17.1) in the AP. We are also given that there are 14 terms in the sequence. Let's use the given information to find the common difference.
We know that the first term (a₁) is 6.7, and the last term (aₙ) is 17.1, where n is the number of terms.
Using the formula mentioned earlier:
aₙ = a₁ + (n - 1) * d
Substituting the given values:
17.1 = 6.7 + (14 - 1) * d
Simplifying the equation:
17.1 = 6.7 + 13d
Rearranging the equation to isolate the common difference (d):
17.1 - 6.7 = 13d
10.4 = 13d
Finally, to find the value of d (common difference):
d = 10.4 / 13
Therefore, the common difference in this arithmetic progression is approximately 0.8.