The fourth term of an arithemetic progression is 13 and the tenth term is 31 find it first tern and it common difference
a+3d = 13
a+9d = 31
subtract them
6d = 18
d = 3
back into the 1st equation:
a + 9 = 13
a= 4
To find the first term (a) and the common difference (d) of an arithmetic progression, we can use the formulas:
n-th term (Tn) = a + (n-1)d,
where Tn represents the n-th term, a is the first term, n is the position of the term in the sequence, and d is the common difference.
Given that the fourth term (T4) is 13 and the tenth term (T10) is 31, we can write two equations:
Equation 1: T4 = a + (4-1)d = 13
Equation 2: T10 = a + (10-1)d = 31
Let's solve these two equations simultaneously to find the first term (a) and the common difference (d).
From Equation 1:
a + 3d = 13
From Equation 2:
a + 9d = 31
To get the value of 'a', let's eliminate 'd' by multiplying Equation 1 by 3 and subtracting it from Equation 2:
3(a + 9d) - 3(a + 3d) = 31 - 13
3a + 27d - 3a - 9d = 18
18d = 18
d = 1
Now, substitute the value of d back into Equation 1:
a + 3(1) = 13
a + 3 = 13
a = 13 - 3
a = 10
Hence, the first term (a) is 10 and the common difference (d) is 1 in the given arithmetic progression.