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

1 answer

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.