In an arithmetic progression the sum of the first 10 terms is 50 and the 5th is three times the second term. Find the fisrt term and the common difference.
To find the first term and common difference of the arithmetic progression, we can use the given information.
Let's denote the first term as "a" and the common difference as "d".
We're given that the sum of the first 10 terms is 50, which means:
S10 = 10/2 * (2a + (10-1)d) = 50
Simplifying this equation, we have:
5(2a + 9d) = 50
2a + 9d = 10 (equation 1)
We're also given that the 5th term is three times the second term:
a + 4d = 3(a + d) (equation 2)
Now we can solve these two equations simultaneously to find the values of "a" and "d".
First, let's expand equation 2:
a + 4d = 3a + 3d
4d - 3d = 3a - a
d = 2a (equation 3)
Now substitute equation 3 into equation 1:
2a + 9(2a) = 10
2a + 18a = 10
20a = 10
a = 10/20
a = 1/2 (first term)
Substitute the value of "a" into equation 3 to find the common difference ("d"):
d = 2(1/2)
d = 1 (common difference)
Therefore, the first term (a) is 1/2 and the common difference (d) is 1.
To solve this problem, we need to use the formulas for the sum of an arithmetic progression and the nth term of an arithmetic progression.
Let's start by finding the first term, denoted as 'a', and the common difference, denoted as 'd', of the arithmetic progression.
Given:
The sum of the first 10 terms is 50:
We can use the formula for the sum of an arithmetic progression:
Sum of n terms = (n/2) * (2a + (n - 1)d)
Substituting the given values:
50 = (10/2) * (2a + (10 - 1)d)
50 = 5(2a + 9d)
10 = 2a + 9d --- (Equation 1)
The 5th term is three times the second term:
The nth term of an arithmetic progression can be calculated using the formula:
an = a + (n - 1)d
Substituting n = 5:
a5 = a + (5 - 1)d
a5 = a + 4d
Given that a5 = 3a (three times the second term), we can substitute this into the equation:
3a = a + 4d --- (Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) that we can solve simultaneously.
First, let's solve Equation 2 for a:
3a = a + 4d
2a = 4d
a = 2d --- (Equation 3)
Substitute Equation 3 into Equation 1:
10 = 2a + 9d
10 = 2(2d) + 9d
10 = 4d + 9d
10 = 13d
d = 10/13
Substitute d = 10/13 back into Equation 3 to find a:
a = 2d
a = 2(10/13)
a = 20/13
Therefore, the first term, a, is 20/13 and the common difference, d, is 10/13.