the first term of an arithmetic progression is 3 and the fifth term is is 9.find the number of terms in the progression if the last term is 81
Isn't your school subject math?
Thank you
It is math
Well, let me put on my arithmetic clown shoes and juggle some numbers for you!
To find the number of terms in the arithmetic progression, we need to figure out the common difference first. The common difference is the constant amount that is added or subtracted to get from one term to the next.
To find the common difference, we can subtract the first term from the fifth term:
9 - 3 = 6
So the common difference is 6. Now we can determine how many terms we have based on the last term of 81.
To get the number of terms, we can use the formula:
n = (last term - first term) / common difference + 1
Plugging in our values:
n = (81 - 3) / 6 + 1
That simplifies to:
n = 78 / 6 + 1
Now, let's divide 78 by 6:
n = 13 + 1
Drumroll please! The number of terms in the progression is 14. Ta-da!
To find the number of terms in the arithmetic progression, we need to determine the common difference first. The common difference (d) is the constant value added to each term to obtain the next term in the sequence.
Given that the first term (a₁) is 3 and the fifth term (a₅) is 9, we can find the common difference (d) using the formula:
d = (a₅ - a₁) / 4
= (9 - 3) / 4
= 6 / 4
= 1.5
Now that we know the common difference is 1.5, we can find the number of terms (n) in the progression. Since the last term (aₙ) is 81, we can use the formula for the nth term of an arithmetic progression:
aₙ = a₁ + (n - 1) * d
Substituting the values we have:
81 = 3 + (n - 1) * 1.5
Simplifying the equation:
81 - 3 = (n - 1) * 1.5
78 = 1.5 * (n - 1)
52 = n - 1
n = 53
Therefore, the number of terms in the arithmetic progression is 53.
well, since
T5-T1 = 4d, d=3/2
we know a=3, so
Sn = n/2 (2*3+(n-1)(3/2))
3/4 (n^2+3n) = 81
n^2+3n-108 = 0
(n+12)(n-9) = 0
n=9
and the sequence is
3 4.5 6 7.5 9 10.5 12 13.5 15
The sum is 81