The sum of the first 12terms of an arithmetic progression is 135 while the sum of the next 8 terms is 210. Find the common difference
old how did you get 345
please how did you get 345
OK I see now . sorry I asked
To find the common difference of an arithmetic progression, we can use the formulas for the sum of the terms of an arithmetic progression.
The sum of the first n terms of an arithmetic progression is given by:
Sn = (n/2)(2a1 + (n-1)d),
where Sn is the sum of the first n terms, a1 is the first term, d is the common difference, and n is the number of terms.
In this case, we are given that the sum of the first 12 terms is 135, so we can write the following equation:
135 = (12/2)(2a1 + 11d).
Simplifying this equation, we have:
135 = 6(2a1 + 11d),
135 = 12a1 + 66d.
Similarly, we are given that the sum of the next 8 terms (terms 13 to 20) is 210, so we can write the following equation:
210 = (8/2)(2a1 + (8-1)d).
Simplifying this equation, we have:
210 = 4(2a1 + 7d),
210 = 8a1 + 28d.
Now, we have a system of two equations with two variables (a1 and d). We can solve this system of equations to find the common difference (d).
First, we can subtract equation 2 from equation 1:
(135 - 210) = (12a1 + 66d) - (8a1 + 28d),
-75 = 4a1 + 38d.
Next, we can solve this equation for a1:
-75 = 4a1 + 38d,
-75 - 38d = 4a1,
-19 - 9.5d = a1.
Now, we substitute this value of a1 back into equation 1:
135 = 12a1 + 66d,
135 = 12(-19 - 9.5d) + 66d,
135 = -228 - 114d + 66d,
135 = -228 - 48d,
363 = -48d,
d = -7.5625.
Therefore, the common difference of the arithmetic progression is -7.5625.
We are told that
S12 = 135
S20-S12 = 210
S20 - 135 = 210
S20 = 345
Now, we know that
S20 = 20/2 (2a+19d)
S12 = 12/2 (2a+11d)
So, we have
10(2a+19d) = 345
6(2a+11d) = 135
Solve that for d.