An arithmetic progression of 41 terms is such that the sum of the first five terms is 560 and the sum of the last five terms is -250 find
a) The five terms and common difference.
b)The last term
c)Sum of the arithmetic progression
a) To find the five terms and common difference, we can use the formula for the sum of an arithmetic progression:
S = (n/2)(2a + (n-1)d),
where S is the sum, n is the number of terms, a is the first term, and d is the common difference.
We are given S = 560 for the sum of the first five terms, so we have:
560 = (5/2)(2a + 4d).
Simplifying this equation gives:
560 = 5a + 10d.
We are also given S = -250 for the sum of the last five terms, so we have:
-250 = (5/2)(2a + 36d).
Simplifying this equation gives:
-500 = 5a + 90d.
Now we have a system of two equations with two unknowns:
5a + 10d = 560,
5a + 90d = -500.
Subtracting the first equation from the second gives:
80d = -1060,
so d = -1060/80 = -13.25.
Substituting this value of d into one of the equations, we can solve for a:
5a + 10(-13.25) = 560,
5a - 132.5 = 560,
5a = 692.5,
a = 138.5.
Therefore, the five terms are:
a = 138.5,
a + d = 125.25,
a + 2d = 112,
a + 3d = 98.75,
a + 4d = 85.5,
and the common difference is d = -13.25.
b) To find the last term, we can use the formula for the nth term of an arithmetic progression:
a_n = a + (n-1)d.
Since there are 41 terms, the last term is:
a_41 = 138.5 + (41-1)(-13.25),
a_41 = 138.5 + 40(-13.25),
a_41 = 138.5 - 530,
a_41 = -391.5.
Therefore, the last term is -391.5.
c) To find the sum of the arithmetic progression, we can use the formula:
Sn = (n/2)(a + a_n).
Substituting the values we found:
S41 = (41/2)(138.5 - 391.5),
S41 = (20.5)(-253),
S41 = -5186.5.
Therefore, the sum of the arithmetic progression is -5186.5.