The sum of the 3rd and the 7th term of an A.P. is 30 and the sum of the 5th…

Let the first term be a and the common difference be d.

1. Write the given sums in terms of a and d:

   a3 + a7 = (a + 2d) + (a + 6d) = 2a + 8d = 30.

   a5 + a9 = (a + 4d) + (a + 8d) = 2a + 12d = 56.

2. Subtract the first equation from the second to eliminate a:

   (2a + 12d) - (2a + 8d) = 56 - 30 ⇒ 4d = 26 ⇒ d = 6.5.

3. Substitute d = 6.5 into 2a + 8d = 30 to find a:

   2a + 8(6.5) = 30 ⇒ 2a + 52 = 30 ⇒ 2a = -22 ⇒ a = -11.

4. Compute the required sum a4 + a8:

   a4 + a8 = (a + 3d) + (a + 7d) = 2a + 10d.

   Plugging in a = -11 and d = 6.5 gives 2(-11) + 10(6.5) = -22 + 65 = 43.

Therefore the sum of the 4th and 8th terms is 43.