A) In an arithmetic progression the sum of the first ten terms is 50 and the fifth term is three times the second term
(1) Calculate the first term
(2)Calculate the sum of the 20 terms
Just translate into Math using the definitions of AP
"the fifth term is three times the second term " ---> a+4d = 3(a+d)
a+4d = 3a+3d
d = 2a **
"the sum of the first ten terms is 50"
--> 5(2a+9d) = 50
2a+9d=10 ***
sub ** into ***
2a+9(2a) = 10
20a=10
a = 1/2
d = 1
first term is 1/2
sum(20)
= 10(2(1/2) + 19(1)) = 200
To solve this problem, we will use the formulas for the sum of an arithmetic progression and the nth term of an arithmetic progression.
Let's denote the first term of the arithmetic progression as "a" and the common difference as "d".
1) Calculate the first term:
We are given that the sum of the first 10 terms is 50. The formula for the sum of an arithmetic progression is:
Sn = (n/2)(2a + (n-1)d),
where Sn is the sum of the first n terms.
Substituting the given values, we get:
50 = (10/2)(2a + (10-1)d)
50 = 5(2a + 9d)
10a + 45d = 50
We are also given that the fifth term (5th term) is three times the second term (2nd term):
a + 4d = 3(a + d)
a + 4d = 3a + 3d
4d - 3d = 3a - a
d = 2a
Substituting this value of "d" into the equation 10a + 45d = 50, we have:
10a + 45(2a) = 50
10a + 90a = 50
100a = 50
a = 50/100
a = 0.5
Therefore, the first term of the arithmetic progression is 0.5.
2) Calculate the sum of the 20 terms:
The formula for the sum of the first n terms of an arithmetic progression is:
Sn = (n/2)(2a + (n-1)d),
where Sn is the sum of the first n terms.
We want to find the sum of the first 20 terms (S20). Substituting the given values, we have:
S20 = (20/2)(2 * 0.5 + (20-1)d)
S20 = 10(1 + 19d)
S20 = 10(1 + 19 * 2 * 0.5) [since d = 2a]
S20 = 10(1 + 19) [simplifying]
S20 = 10(20)
S20 = 200
Therefore, the sum of the first 20 terms of the arithmetic progression is 200.
To solve this problem, we can use the formulas for the sum of an arithmetic progression and the formula for the nth term of an arithmetic progression.
The sum of the first n terms of an arithmetic progression is given by the formula:
Sn = (n/2)(2a + (n-1)d),
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
We are given that the sum of the first ten terms is 50. So, we have:
(10/2)(2a + (10-1)d) = 50.
We are also given that the fifth term is three times the second term. So, we have:
a + 4d = 3(a + d).
Now, let's solve these equations to find the values of a and d.
(1) Calculate the first term (a):
From the equation a + 4d = 3(a + d), we can simplify it as follows:
a + 4d = 3a + 3d,
2d = 2a,
a = d.
Substituting this result into the equation (10/2)(2a + (10-1)d) = 50, we get:
(10/2)(2a + (10-1)a) = 50,
5a(2 + 9) = 50,
5a(11) = 50,
55a = 50,
a = 50/55,
a = 10/11.
So, the first term (a) is 10/11.
(2) Calculate the sum of the 20 terms (S20):
Now that we know the value of a, we can calculate the sum of the 20 terms using the formula Sn = (n/2)(2a + (n-1)d).
Substituting the values into the formula, we have:
S20 = (20/2)(2(10/11) + (20-1)(10/11)),
S20 = 10(20/11 + 19(10/11)),
S20 = 10(20/11 + 190/11),
S20 = 10(210/11),
S20 = (2100/11).
Therefore, the sum of the 20 terms is 2100/11.