An arithmetic progression has 5 terms that have a sum of 200. The sum of the last three terms added to 6 times the sum of the first two terms equals zero. What is the first term?
a1 + a2 + a3 + a4 + a5 = 200
6 (a1 + a2) + a3 + a4 + a5 = 0
a1 = a1
a2 = a1 + m
a3 = a1 + 2m
a4 = a1 + 3m
a5 = a1 + 4m
so
5 a1 + 10m = 200
6(2 a1+ m) + 3 a1 + 9m = 0
a1 + 2 m = 40
15 a1 + 15 m = 0
m = -a1
a1 - 2a1 = 40
a1 = -40
m = 40
Why did the arithmetic progression go to therapy? Because it had trouble summing up its feelings! Alright, let's solve this.
Let's call the first term of the arithmetic progression "a", and let "d" be the common difference between each term.
We are given that the sum of the five terms is 200, so we can write the equation as:
(5/2)[2a + 4d] = 200
Now, we're also given that the sum of the last three terms added to 6 times the sum of the first two terms equals zero. We can write another equation for this information:
[(a + 2d) + (a + 3d) + (a + 4d)] + 6[(a + (a + d))] = 0
Now it's time to put on our solving hats and find the values of "a" and "d". Let's simplify these equations:
1. From the first equation:
(5/2)[2a + 4d] = 200
10a + 20d = 200
Divide both sides by 10:
a + 2d = 20 ---------- (Equation 1)
2. From the second equation:
[(a + 2d) + (a + 3d) + (a + 4d)] + 6[(a + (a + d))] = 0
Combine like terms:
4a + 10d = -3a - 3d
7a + 13d = 0 ---------- (Equation 2)
Now, we have a system of two equations. Let's solve it using substitution or elimination to find the values of "a" and "d". Once we have "a", we have our answer!
Let's solve this step-by-step.
Step 1: Determine the sum of the first five terms of the arithmetic progression.
Arithmetic progressions have the formula Sn = (n/2)(2a + (n-1)d), where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
In this case, the sum of the first five terms is 200, so we have: Sn = 200.
Step 2: Find the sum of the last three terms.
According to the problem, the sum of the last three terms added to 6 times the sum of the first two terms equals zero. This can be expressed as: (a + 4d) + (a + 3d) + (a + 2d) + 6(a + a + d) = 0.
Step 3: Simplify the equation from step 2.
By combining like terms, we get: 3a + 9d = 0.
Step 4: Set up a system of equations using the equations from steps 1 and 3.
We have two equations:
Sn = 200 ...(Equation 1)
3a + 9d = 0 ...(Equation 2)
Step 5: Solve the system of equations.
From Equation 1, we have Sn = (5/2)(2a + 4d) = 200.
Simplifying Equation 1, we get: 5a + 10d = 200.
Now we have a system of equations:
5a + 10d = 200 ...(Equation 3)
3a + 9d = 0 ...(Equation 2)
Step 6: Solve the system of equations for a and d.
By multiplying Equation 2 by 5 and Equation 3 by 3, we can eliminate d. This gives us:
15a + 45d = 0 ...(Equation 4)
15a + 30d = 600 ...(Equation 5)
Subtracting Equation 4 from Equation 5, we get:
15a + 30d - 15a - 45d = 600 - 0
-15d = -600
d = (-600)/(-15)
d = 40
Substituting the value of d into Equation 2, we get:
3a + 9(40) = 0
3a + 360 = 0
3a = -360
a = (-360)/3
a = -120
Therefore, the first term of the arithmetic progression is -120.
To solve this problem, we need to find the first term of the arithmetic progression. Let's break down the information given step by step:
Step 1: Determine the sum of all five terms.
The sum of an arithmetic progression is given by the formula:
S = (n/2)(2a + (n-1)d)
where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
In our case, the sum of the five terms is given as 200.
So, we can set up the equation:
200 = (5/2)(2a + 4d)
Step 2: Find the sum of the last three terms.
Let's denote the last three terms as t1, t2, and t3.
The sum of the last three terms is given by:
Sum of the last three terms = t1 + t2 + t3 = 3t3
Since it is an arithmetic progression, the common difference (d) remains the same.
So, the last term (t3) can be represented as:
t3 = a + 4d
Step 3: Find the sum of the first two terms.
The sum of the first two terms is given by:
Sum of the first two terms = a + (a + d) = 2a + d
Step 4: Use the given equation to set up another equation.
We are given that the sum of the last three terms added to 6 times the sum of the first two terms equals zero.
So, we can set up the equation:
3t3 + 6(2a + d) = 0
Substituting the value of t3 from Step 2:
3(a + 4d) + 6(2a + d) = 0
Step 5: Solve the two equations.
Now we have two equations:
200 = (5/2)(2a + 4d) -- Equation 1
3(a + 4d) + 6(2a + d) = 0 -- Equation 2
Solving Equation 2, we get:
3a + 12d + 12a + 6d = 0
15a + 18d = 0
Divide both sides by 3:
5a + 6d = 0 -- Equation 3
Now, we have equations 1 and 3 with two variables (a and d). We can solve them simultaneously to find their values.
Step 6: Solve the simultaneous equations.
Equation 1: 200 = (5/2)(2a + 4d)
By simplifying, we get:
40 = 2a + 4d
Divide both sides by 2:
20 = a + 2d -- Equation 4
Equation 3: 5a + 6d = 0
Multiplying both sides by 2, we get:
10a + 12d = 0 -- Equation 5
Now, we can subtract Equation 5 from Equation 4 to eliminate the variable d:
(20 - 10a) - (10a + 12d) = 0
20 - 10a - 10a - 12d = 0
20 - 20a - 12d = 0
Divide both sides by -12:
-12d = 20a - 20
12d = 20 - 20a
d = (20 - 20a) / 12
d = (5 - 5a/3) -- Equation 6
Substitute Equation 6 into Equation 4:
20 = a + 2d
20 = a + 2((5 - 5a/3))
20 = a + 10 - (10a/3)
60 = 3a + 30 - 10a
60 - 30 = -7a
30 = -7a
a = -30/7
Therefore, the first term (a) is -30/7 or approximately -4.29.