Answer:
(a) To find the value of a and d, we use the formula for the sum of first n terms of the arithmetic series A which is given by:Sn = n/2[2a + (n-1)d]We are also given that Sn = 15 + 2n. So we can equate these two expressions to get:15 + 2n = n/2[2a + (n-1)d]Multiplying both sides by 2 and simplifying, we get:30 + 4n = n[2a + (n-1)d]Expanding the brackets and simplifying, we get:2an + nd - d + 30 = 2n^2Rearranging terms, we get:2a = 2n^2 - nd + d - 30Now we also know that the first term of the series A is a. So we can substitute this value of a in the formula above to get:a = (2n^2 - nd + d - 30)/2Simplifying, we get:a = n^2 - (n-1)d - 15Therefore, we have found the values of a and d in terms of n. (b) To find the 20th term of A, we use the formula for the nth term of an arithmetic series which is given by:an = a + (n-1)dSubstituting the value of a and d that we found in part (a) we get:a20 = (20^2 - 19d - 15) + 19dSimplifying, we get:a20 = 391 - dTherefore, the 20th term of A is given by a20 = 391 - d.(c) Given that S2p - 2Sp = 1 + S(p-1), we can use the formula for the sum of first n terms of an arithmetic series which we used in part (a) to get:2p/2[2a + (2p-1)d] - 2p/2[2a + (p-1)d] = 1 + p/2[2a + (p-2)d]Simplifying, we get:2apd = d(p^2 - 3p + 2)Dividing both sides by d and simplifying, we get:2ap = p^2 - 3p + 2Rearranging terms, we get:p^2 - 3p + (2-2ap) = 0This is a quadratic equation with coefficients a=1, b=-3, and c=2-2ap. We can use the quadratic formula to solve for p:p = [3 ± sqrt(9 - 4(1)(2-2ap))]/2Simplifying, we get:p = [3 ± sqrt(4ap + 1)]/2Therefore, we have found the value of p in terms of a.