Question 998264

nth term = 3-2n

t1= 3-2=1

t2= 3-4=-1

t3=3-6 = -3

The AP is 1,-1,-3 .........


here a = 1, d= -2 , n=15

Sn = {{{ (n/2)(2a+(n-1)d)}}}

S 15 = {{{ (15/2)(2(1)+(15-1)*(-2))}}}

S15 = {{{(15/2) * -26}}}

= -13*15

=-195