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

Answer 1

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.