The 5th term of an arithmetic progression is 82 and the 12th term is 103.Fsum of the first 21 terms
To find the sum of the first 21 terms of an arithmetic progression, we need to find the common difference and the first term of the sequence.
Let's start by finding the common difference (d). We are given that the 5th term is 82 and the 12th term is 103.
Let's use the formula for the nth term of an arithmetic progression:
An = A1 + (n-1)d
For the 5th term (n = 5):
82 = A1 + 4d
For the 12th term (n = 12):
103 = A1 + 11d
Now we have a system of two equations:
82 = A1 + 4d
103 = A1 + 11d
Let's solve this system of equations:
Multiplying the first equation by 11 and the second equation by 4 gives us:
902 = 11A1 + 44d
412 = 4A1 + 44d
Subtracting these two equations gives us:
902 - 412 = 11A1 - 4A1
490 = 7A1
Dividing both sides by 7 gives us:
A1 = 70
Now, we can substitute the value of A1 into either of the original equations to find the common difference:
82 = 70 + 4d
4d = 82 - 70
4d = 12
d = 12/4
d = 3
So, the common difference (d) is 3 and the first term (A1) is 70.
Now we can find the sum of the first 21 terms of the arithmetic progression using the formula:
Sn = n/2 * (A1 + An)
Sn = 21/2 * (70 + (70 + (21-1)*3))
Sn = 21/2 * (70 + (70 + 20*3))
Sn = 21/2 * (70 + (70 + 60))
Sn = 21/2 * (70 + 130)
Sn = 21/2 * 200
Sn = 21 * 100
Sn = 2100
So, the sum of the first 21 terms of the arithmetic progression is 2100.