The sum of third and ninth terms of arithmetic series is 20 and the difference between the twelfth and fourth term is 32.determine the value of the first term and the constant difference.
using the definitions:
a+2d + a+8d = 20
2a + 10d = 20
a + 5d = 10 **
a+11d - (a+3d) = 32
a+11d - a - 3d = 32
8d = 32
d = 4
sub into **
a + 20 = 10
a = -10
state your conclusion.
To solve this problem, we need to use the formulas for the nth term of an arithmetic series and the formula for the difference between two terms.
Let's denote the first term by 'a' and the common difference by 'd'.
We are given two conditions:
1. The sum of the third and ninth terms is 20.
The Formula for the nth term is: tn = a + (n-1)d
Using this formula, we can write the third term as t3 = a + 2d and the ninth term as t9 = a + 8d.
According to the given condition, the sum of t3 and t9 is 20, so we have the equation:
t3 + t9 = 20
Substituting the values of t3 and t9, we get:
(a + 2d) + (a + 8d) = 20
Simplifying the equation:
2a + 10d = 20 ----(equation 1)
2. The difference between the twelfth and fourth terms is 32.
Using the formula for the difference between two terms, we can write the equation as:
t12 - t4 = 32
Substituting the values of t12 and t4 using the nth term formula, we get:
(a + 11d) - (a + 3d) = 32
Simplifying the equation:
8d = 32 ----(equation 2)
Now, we have a system of two equations (equation 1 and equation 2) with two variables (a and d).
Using equation 2, we can solve for d:
8d = 32
d = 32/8
d = 4
Substituting the value of d back into equation 1, we can solve for a:
2a + 10(4) = 20
2a + 40 = 20
2a = 20 - 40
2a = -20
a = -20/2
a = -10
Therefore, the first term (a) is -10 and the common difference (d) is 4.