Show that a1, a2, ..., an form an AP where an is defined as below: (i) an = 3 + 4n (ii) an = 9 - 5n. Also find the sum of the first 15 terms in each case.

(i) an = 3 + 4n
given
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 11
a3 = 3 + 4(3) = 15
∴d = a2 - a1 = 11 - 7 = 4
Here a = 7, d = 4 and n = 15
By using Sn = n/2[2a + (n - 1)d] we have,
S15 = 15/2[2 × 7 + (15 - 1)4] = 15/2(14 + 56) = 15/2 × 70 = 525
(ii) an = 9 - 5n
given
a1 = 9 - 5(1) = 9 - 5 = 4
a2 = 9 - 5(2) = 9 - 10 = -1
a3 = 9 - 5(3) = 9 - 15 = -6
∴d = a2 - a1 = -6 - (-1) = -5
Here a = 4, d = -5 and n = 15
By using Sn = n/2[2a + (n - 1)d] we have,
S15 = 15/2[2 × 4 + (15 - 1)(-5)] = 15/2(8 - 70) = 15/2 × (-62) = -465