it is given that the sum of the seventh and tenth term of an arithmetic sequence is 12 and term 13 is equal to "-30"
determine the sum of the first 100 terms in the sequence.
Since we know that the 13th term is -30, use the explicit formula for an arithmetic sequence, plugging in n=13 and solving for a1:
a(n) = a1+(n-1)d
a(13) = a1+(13-1)d
-30 = a1+12d
a1 = -30-12d
Since the 7th and 10th terms add up to 12, do the same thing:
a1+(7-1)d + a1+(10-1)d = 12
2(a1) + 6d + 9d = 12
2(a1) + 15d = 12
a1 + (15/2)d = 6
a1 = 6 - (15/2)d
Now, set the equations equal to each other to find d, the common difference:
-30 - 12d = 6 - (15/2)d
-30 = 6 + (9/2)d
-36 = (9/2)d
-8 = d
Lastly, find the value of a1:
a1 = 6 - (15/2)(-8)
a1 = 6 + 120/2
a1 = 6 + 60
a1 = 66
Therefore, your arithmetic sequence is a(n)=66+(n-1)(-8)
Using this information, we also know that the equation for an arithmetic series is S(n) = n[(a1 + a(n))/2], so we need to find what a(100) is and then we can determine S(100):
a(100) = 66 + (100-1)(-8)
a(100) = 66 + (99)(-8)
a(100) = 66 - 792
a(100) = -726
Therefore, we can determine S(100), which represents the sum of the first 100 terms in the sequence:
S(100) = 100[(66+a(100))/2]
S(100) = 100[(66-726)/2]
S(100) = 100[-660/2]
S(100) = 100(-330)
S(100) = -33000
In conclusion, the sum of the first 100 terms in the sequence is -33,000.
Hope this explanation helped!
thank you, for all the help
a+6d + a+9d = 12
a+12 = -30
solve for a and d, and then you want
S100 = 100/2 (2a+99d)
To find the sum of the first 100 terms of an arithmetic sequence, we need to determine the common difference (d) and the first term (a₁).
Given:
T₇ + T₁₀ = 12 -- (Equation 1)
T₁₃ = -30 -- (Equation 2)
Step 1: Calculating the common difference (d)
We know that every term in an arithmetic sequence can be expressed as:
Tₙ = a₁ + (n - 1) * d,
where:
Tₙ is the nth term,
a₁ is the first term,
n is the position of the term, and
d is the common difference.
Substitute n = 7 into this equation:
T₇ = a₁ + (7 - 1) * d
= a₁ + 6d
Similarly, substitute n = 10:
T₁₀ = a₁ + (10 - 1) * d
= a₁ + 9d
Now, we can rewrite Equation 1 using these expressions:
T₇ + T₁₀ = 12
a₁ + 6d + a₁ + 9d = 12
2a₁ + 15d = 12
Step 2: Calculating the first term (a₁)
To solve for a₁, we need a second equation. For this, we can use Equation 2:
T₁₃ = a₁ + (13 - 1) * d
= a₁ + 12d
Substituting T₁₃ = -30 into this equation:
a₁ + 12d = -30
Step 3: Solving the system of equations
Now we have two equations:
2a₁ + 15d = 12 -- (Equation 3)
a₁ + 12d = -30 -- (Equation 4)
To solve this system, we can use the method of substitution or elimination.
Let's use the method of substitution:
From Equation 4, isolate a₁:
a₁ = -30 - 12d
Substitute this value into Equation 3:
2(-30 - 12d) + 15d = 12
-60 - 24d + 15d = 12
-60 - 9d = 12
-9d = 12 + 60
-9d = 72
d = -72 / -9
d = 8
Now that we have the common difference (d = 8), we can substitute it back into Equation 4 to find the first term (a₁):
a₁ + 12(8) = -30
a₁ = -30 - 96
a₁ = -126
Step 4: Calculating the sum of the first 100 terms
To find the sum of the first 100 terms (S₉₉), we use the arithmetic series formula:
Sₙ = [n/2] * (2a₁ + (n - 1) * d)
Substitute n = 100, a₁ = -126, and d = 8:
S₁₀₀ = [100/2] * (2(-126) + (100 - 1) * 8)
= 50 * (-252 + 99 * 8)
= 50 * (-252 + 792)
= 50 * 540
= 27,000
Therefore, the sum of the first 100 terms in the arithmetic sequence is 27,000.