In an arithmetic progression the 5th term is six times the 1st term and the sum of the first six terms is 99. Find the 1st term and the common difference.
Solution
To solve this problem, let's assign variables to the given information. Let's say the first term of the arithmetic progression is "a" and the common difference is "d".
We know that the 5th term is six times the 1st term, so we can write this as:
a + 4d = 6a
Now let's find the sum of the first six terms of the arithmetic progression using the formula:
Sum of n terms = (n/2) * (2a + (n - 1) * d)
For the first six terms, the sum is 99, so we have:
(6/2) * (2a + (6 - 1) * d) = 99
3 * (2a + 5d) = 99
6a + 15d = 99
Now we have a system of two equations:
1) a + 4d = 6a
2) 6a + 15d = 99
To solve this system, we can use substitution or elimination method.
Let's use the substitution method:
Rearrange equation 1 to solve for "a" in terms of "d":
a = 4d/(6-1)
a = 4d/5
Substitute this value of "a" into equation 2:
6(4d/5) + 15d = 99
Multiply through by 5 to get rid of fractions:
24d + 75d = 495
99d = 495
d = 495/99
d = 5
Now that we know the common difference "d" is 5, we can substitute it back into equation 1 to find the first term "a":
a + 4(5) = 6a
a + 20 = 6a
20 = 6a - a
20 = 5a
a = 20/5
a = 4
Therefore, the first term is 4 and the common difference is 5.