Question

The sum of first n ap is given by sn=3n2-4n determine the ap and the 12 term

Answer 1

Sn = 3n²-4n

Sn-S(n-1) = An

A1 = 3*1²-4*1

= -1

A2 = 3*2²-4*2-(3*1²-4*1)

= 12-4-(-1)

= 8+1

= 9

∴ d= a2-a1

d = 9-(-1)

d = 10

Therefore AP is :

-1,9,19,29,39........................

A12 = a+11d

= -1+11*10

= 110-1

A12 = 109