The sum of 10th term of an AP is 30 and it's 12th term is 36 .find the first term and the common difference
Let's denote the first term of the AP as "a" and the common difference as "d".
The formula for the nth term of an AP is given by:
a_n = a + (n-1)d
We are given that the sum of the 10th term is 30, so we can set up the equation:
a + (a + d) + (a + 2d) + ... + (a + 9d) = 30
We are also given that the 12th term is 36, so we can set up another equation:
a + 11d = 36
Using the formula for the sum of an AP, the sum of the 10th term can be expressed as:
S_10 = (10/2)(2a + (10-1)d) = 30
We can simplify this equation to:
5(2a + 9d) = 30
2a + 9d = 6
Now we have two equations with two variables:
2a + 9d = 6
a + 11d = 36
Multiplying the first equation by 2 and subtracting the second equation from it, we get:
4a + 18d - (a + 11d) = 12 - 36
3a = -24
a = -8
Substituting the value of "a" into the second equation, we can solve for "d":
-8 + 11d = 36
11d = 44
d = 4
Therefore, the first term is -8 and the common difference is 4.