The sum of the first 10 terms of an arithmetic series is 145 and the sum of its fourth and ninth term is five times the third the..determine the first term and constant difference
To determine the first term and constant difference of the arithmetic series, we can use the given information about the sum of the first 10 terms and the relationship between the fourth, ninth, and third terms.
Let's break down the problem step by step:
Step 1: Find the formula for the sum of an arithmetic series.
The formula for the sum of an arithmetic series is given by:
S = (n/2) * (2a + (n-1)d)
where:
S is the sum of the series
n is the number of terms in the series
a is the first term of the series
d is the constant difference between the terms
Step 2: Use the given sum of the first 10 terms to find the equation.
We are told that the sum of the first 10 terms is 145. Substituting this information into the arithmetic series formula:
145 = (10/2) * (2a + (10 - 1)d)
Simplifying further:
145 = 5(2a + 9d)
29 = 2a + 9d -- Equation 1
Step 3: Use the given relationship between the fourth, ninth, and third terms to find the equation.
We are told that the sum of the fourth and ninth term is five times the third term. Let's represent these terms as:
Third term: a + 2d -- Equation 2
Fourth term: a + 3d
Ninth term: a + 8d
According to the given information:
(a + 3d) + (a + 8d) = 5(a + 2d)
Simplifying the equation:
2a + 11d = 5a + 10d
-3a = -d
d = 3a -- Equation 3
Step 4: Solve the system of equations.
Now, we can substitute Equation 3 into Equation 1 to solve for a and d:
29 = 2a + 9(3a)
29 = 2a + 27a
29 = 29a
a = 1
Using this value of a, we can find d:
d = 3a
d = 3(1)
d = 3
Therefore, the first term of the arithmetic series is 1 and the constant difference is 3.
10/2(2a+9d) = 145
a+3d + a+8d = 5(a+2d)
a=1
d=3
To find the first term (a) and the common difference (d) of an arithmetic series, we can use the given information.
Let's first find the sum of the arithmetic series using the formula:
Sn = n/2 * (2a + (n-1)d)
where Sn represents the sum of the first n terms, a is the first term, and d is the common difference.
From the given information:
1) Sum of the first 10 terms = 145:
145 = 10/2 * (2a + (10-1)d)
145 = 5 * (2a + 9d)
29 = 2a + 9d ...........(Equation 1)
2) Sum of the fourth and ninth term is five times the third term:
a + 3d + a + 8d = 5(a + 2d)
2a + 11d = 5a + 10d
5d = 3a ...........(Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) that we can solve to find the values of a and d.
To do so, let's isolate a in Equation 2:
3a = 5d
a = 5/3 * d
Substitute this value of a into Equation 1:
29 = 2(5/3 * d) + 9d
29 = 10/3 * d + 9d
87 = 10d + 27d
87 = 37d
Now, solve for d:
d = 87/37
d = 2.3514 (approximately)
Substitute the value of d back into Equation 2 to find a:
a = 5/3 * (2.3514)
a = 3.9189 (approximately)
Therefore, the first term (a) is approximately 3.9189 and the common difference (d) is approximately 2.3514.