If the sixth term of an arithmetic progression is 37 and the sum of the first sixth terms is 147 find the first term
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To find the first term of an arithmetic progression, we need to gather some information.
Let's assume that the first term of the arithmetic progression is denoted by 'a' and the common difference is denoted by 'd'.
Given that the sixth term of the arithmetic progression is 37, we can use the formula for the nth term of an arithmetic progression:
an = a + (n-1)d
Substituting n = 6, a6 = 37, and rearranging the formula, we get:
37 = a + 5d (Equation 1)
Next, we are given that the sum of the first six terms of the progression is 147. We can use the formula for the sum of an arithmetic progression:
Sn = (n/2)(2a + (n-1)d)
Substituting n = 6 and Sn = 147, we get:
147 = (6/2)(2a + 5d)
147 = 3(2a + 5d)
49 = 2a + 5d (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two variables (a and d). We can solve this system to find the values of a and d.
Multiplying Equation 1 by 2, we get:
74 = 2a + 10d (Equation 3)
Subtracting Equation 3 from Equation 2, we eliminate the term '2a', and we obtain:
49 - 74 = 2a + 5d - (2a + 10d)
-25 = -5d
5d = 25
d = 5
Substituting the value of d = 5 into Equation 2, we can find the value of a:
49 = 2a + 5(5)
49 = 2a + 25
24 = 2a
a = 12
Therefore, the first term of the arithmetic progression is 12.