In an arithmetic series, the terms of the series are equally spread out. For example, in 1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3,the last term is 136, and the sum is 1,390, what are the first 3 terms?
Why did the mathematician join the circus? Because he wanted to be in an arithmetic series! Now, let's solve this question together.
We know that the first term of the series is 3 and the last term is 136. To find the common difference, we subtract the first term from the last term: 136 - 3 = 133.
Next, we can use the formula for the sum of an arithmetic series, which is given by: Sn = (n/2)(2a + (n-1)d), where Sn is the sum, a is the first term, d is the common difference, and n is the number of terms.
Plugging in the given values, we have: 1390 = (n/2)(2 * 3 + (n-1) * 133)
Simplifying further, we get: 1390 = (n/2)(6 + 133n - 133)
Multiply everything out: 1390 = (n/2)(133n - 127)
Now, let's solve for n. We can divide both sides by (n/2): 1390 = 133n - 127
Adding 127 to both sides, we have: 1517 = 133n
And finally, dividing by 133: n ≈ 11.39
Since n must be a whole number, the number of terms in the series is approximately 11.
Now, we can find the second term using the common difference: 3 + 133 = 136
And the third term: 136 + 133 = 269
So, the first three terms of the arithmetic series are 3, 136, and 269.
To find the first three terms of an arithmetic series given the first term, the last term, and the sum, we need to use the formulas for the nth term and the sum of an arithmetic series.
The formula for the nth term of an arithmetic series is:
an = a1 + (n - 1)d
where:
an = nth term
a1 = first term
n = number of terms
d = common difference
The formula for the sum of an arithmetic series is:
Sn = (n/2)(a1 + an)
where:
Sn = sum of the series
We have the following information:
a1 = 3 (first term)
an = 136 (last term)
Sn = 1,390 (sum)
d = ?
Using the formula for the sum, we can find the number of terms (n):
1,390 = (n/2)(3 + 136)
1,390 = (n/2)(139)
1,390*2 = 139n
2,780 = 139n
n = 20
Now that we know the number of terms (n), we can find the common difference (d) using the formula for the nth term:
136 = 3 + (20 - 1)d
136 = 3 + 19d
19d = 136 - 3
19d = 133
d = 133/19
d ≈ 7
Now we can find the first three terms of the arithmetic series:
a1 = 3 (given)
a2 = a1 + d = 3 + 7 = 10
a3 = a2 + d = 10 + 7 = 17
Therefore, the first three terms of the arithmetic series are 3, 10, and 17.
To find the first three terms of an arithmetic series, we need to find the common difference (the difference between consecutive terms) and then use it to calculate the first three terms.
Given that the first term (a₁) is 3, the last term (aₙ) is 136, and the sum (S) is 1,390, let's proceed step by step:
Step 1: Find the number of terms (n) in the series.
We can use the formula for the sum of an arithmetic series: S = (n/2)(a₁ + aₙ)
Substitute the given values: 1,390 = (n/2) (3 + 136)
Simplify: 1,390 = (n/2) (139)
Step 2: Solve for n.
Divide both sides of the equation by 139: (2 * 1,390) / 139 = n
Simplify: n = 20
Step 3: Calculate the common difference (d).
Since consecutive terms in an arithmetic series are equally spread out, the common difference (d) can be found using the formula: d = (aₙ - a₁) / (n - 1)
Substitute the given values: d = (136 - 3) / (20 - 1)
Simplify: d = 133 / 19
Simplify further: d ≈ 7
Step 4: Compute the first three terms.
Using the formula for the nth term of an arithmetic series: aₙ = a₁ + (n - 1) * d
Substitute the known values: a₁ = 3, n = 1, and d ≈ 7
a₁ = 3
a₂ = 3 + (2 - 1) * 7 = 10
a₃ = 3 + (3 - 1) * 7 = 17
Therefore, the first three terms of the arithmetic series are 3, 10, and 17.