Question 1033214
3a-5, a+1, 7-a,..., 109-35a


first term = 3a-5

d= (a+1)-(3a-5) = -2a+6

tn = 109-35a


109-35a = (3a-5)+(n-1)(-2a+6)

{{{(114-38a ) /(-2a+6) = n-1}}}

{{{(38(3-a))/(2(3-a))=n-1}}}

n=20