Find the first three terms of an arithmetic series in which the fifteenth term is –7 and the sum of the first fifteen terms is 210
what is the equation??
To find the first three terms of an arithmetic series, we need to determine the common difference and the first term.
Let's denote the first term as 'a' and the common difference as 'd'.
We know that the fifteenth term is -7, so we can write the equation for the fifteenth term as:
a + (15-1)d = -7
Simplifying this equation, we get:
a + 14d = -7 -- Equation 1
We are also given that the sum of the first fifteen terms is 210. The formula for the sum of 'n' terms in an arithmetic series is:
Sn = (n/2)(2a + (n-1)d)
Substituting n = 15 and Sn = 210 into the formula, we get:
210 = (15/2)(2a + (15-1)d)
210 = (15/2)(2a + 14d)
Dividing both sides by (15/2), we get:
14(2a + 14d) = 210
Simplifying this equation, we get:
4a + 28d = 30 -- Equation 2
We now have two simultaneous equations, Equation 1 and Equation 2, with two unknowns (a and d).
Let's solve these equations simultaneously to find the values of 'a' and 'd':
Multiplying Equation 1 by 2, we have:
2a + 28d = -14
Subtracting this equation from Equation 2, we get:
4a + 28d - (2a + 28d) = 30 - (-14)
2a = 44
a = 22
Now substituting the value of 'a' in Equation 1, we get:
22 + 14d = -7
14d = -29
d = -29/14
So the first term (a) is 22, and the common difference (d) is -29/14.
The first three terms of the arithmetic series can be found by substituting the values of 'a' and 'd' into the formula for the nth term of an arithmetic series:
First term = a = 22
Second term = a + d = 22 + (-29/14) = 22 - (29/14) = 22 - 2.07 ≈ 19.93
Third term = first term + 2*d = 22 + 2*(-29/14) = 22 - (58/14) = 22 - 4.14 ≈ 17.86
Therefore, the first three terms of the arithmetic series are approximately 22, 19.93, and 17.86.
fifteenth term is –7 -----> a+14d = -7 **
the sum of the first fifteen terms is 210
----> (15/2)(2a + 14d) = 210
2a + 14d = 28 ***
subtract ** from ***
a = 35
back in **,
a + 14d = -7
35+14d = -7
d = -3
Now form the first 3 terms