Question

which term of the AP 3,15,27,39 will be 132 more than it's 60th term?

Answer 1

a=3

d=a2-a1=15-3 = 12

we know,

an= a+(n-1)d

a60= a+ (60-1)d

= 3 + 59*12 = 708 + 3= 711

So required term is 711 + 132= 843

an=840=a+(n-1)d

843=3+(n-1)12

840=(n-1)12

n-1=840/12

n-1=70

n=71

result: 71st term