The third tem of an A.P is 25 and the tenth term is -3 . Find the first term and the common difference.

a + d(n-1)

25 = a + d(2)
-3 = a + d(9)
------------ subtract
28 = -7 d
d = -4

25 = a - 8
a = 33

To find the first term and the common difference of an arithmetic progression (A.P.), we can use the formula for the nth term of an A.P.:

\[a_n = a + (n - 1)d\]

where \(a_n\) is the nth term, \(a\) is the first term, \(n\) is the term number, and \(d\) is the common difference.

Given that the third term (\(a_3\)) is 25 and the tenth term (\(a_{10}\)) is -3, we can write the following equations:

\[a_3 = a + 2d = 25\]
\[a_{10} = a + 9d = -3\]

Now, we have a system of two equations with two variables. We can solve these equations simultaneously to find the values of \(a\) and \(d\).

First, we'll multiply the first equation by 9 and the second equation by 2 to eliminate the variable \(a\):

\[9(a + 2d) = 9 \cdot 25\]
\[2(a + 9d) = 2 \cdot (-3)\]

Simplifying these equations gives us:

\[9a + 18d = 225\]
\[2a + 18d = -6\]

Now, subtract the second equation from the first equation to eliminate \(a\):

\[(9a + 18d) - (2a + 18d) = 225 - (-6)\]
\[7a = 231\]

Dividing both sides of the equation by 7 gives us:

\[a = 33\]

Now, substitute the value of \(a\) back into either of the original equations to solve for \(d\). Using the first equation:

\[a + 2d = 25\]
\[33 + 2d = 25\]
\[2d = 25 - 33\]
\[2d = -8\]
\[d = -4\]

So, the first term (\(a\)) is 33 and the common difference (\(d\)) is -4.

To find the first term and the common difference of an arithmetic progression (A.P), we can use the formula for the nth term of an A.P:

An = A + (n - 1)d

where An is the nth term, A is the first term, n is the term number, and d is the common difference.

Given that the third term (A3) is 25 and the tenth term (A10) is -3, we can utilize this information to set up two equations:

A3 = A + (3 - 1)d = 25
A10 = A + (10 - 1)d = -3

Solving these equations simultaneously will help us determine both the first term (A) and the common difference (d):

From equation 1, we have:
A + 2d = 25 [Equation 1]

From equation 2, we have:
A + 9d = -3 [Equation 2]

We can now solve these equations simultaneously to find the values of A and d.