Find the sum of the series

Question

The first term of an Arithmetic Progression is (3x+5) where "x" is a positive integer and the last term of the arithmetic progression is (17x+17) and the common difference is 2.(a) Find the number of terms in terms of x". (b)Show that the sum of the series is divisible by 14 only when "x" is odd.

Answer 1

a1 = 3x+5, last term = 17x +17 d = 2

an = 3x+5 +2(n-10) = 3x -2n -15 = the nth term

sn = (n/2)(a1 + last term = (n/2)(3x+5 +17x +17) = (n/2)(20x +22)

last term = an = 17x+17 = 3x -2n -15

19x +32 = -2n

n = -19x -32)/2 = -9.5x - 16 x has to be even or n = a fraction

x also has to be 2 or greater or n is negative

let x =2, then

a1 = 11, and an = 51

61 = 11 +2(n-1)

25 = n-1

n = 26

17+17x is the 26th term