An arithmetic progression has 2 as its first term. find its common difference if the sum of 33 terms is equal to six times the sum of 13 terms
Sum(33) = 6Sum(13)
(33/2)(2a + 32d) = 6(13/2)(2a + 12d)
but a = 2
33(4 + 32d) = 78(4 + 12d)
sove for d
To solve this problem, let's denote the first term of the arithmetic progression (AP) as 'a' and the common difference as 'd'. We know that the first term, 'a', is 2.
We are given that the sum of 33 terms of the AP is equal to six times the sum of 13 terms. Let's call the sum of the 33 terms S33 and the sum of the 13 terms S13.
Therefore, we have the equation:
S33 = 6 * S13
The sum of an arithmetic progression can be calculated using the formula:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn is the sum of the first n terms,
a is the first term,
n is the number of terms,
and d is the common difference.
Let's find S33 and S13 using this formula.
For S33:
n = 33
a = 2
d = ?
S33 = (33/2)(2(2) + (33-1)d)
S33 = (33/2)(4 + 32d)
S33 = 66 + 528d
For S13:
n = 13
a = 2
d = ?
S13 = (13/2)(2(2) + (13-1)d)
S13 = (13/2)(4 + 12d)
S13 = 26 + 78d
Now, substituting these values into our initial equation:
66 + 528d = 6 (26 + 78d)
Let's simplify and solve for d:
66 + 528d = 156 + 468d
60d = 90
d = 90/60
d = 3/2
Therefore, the common difference of the arithmetic progression is 3/2.