Answers

2016-02-01T20:41:34+05:30
Given AP : 3,15,27,39,.......
a = 3
d = 15 - 3 = 12
Given that a term is 132 more than its 60th term
Here T60 = a+ 59d
= 3+59(12)
= 3+708
= 711
Given that the term is 132 more than its 60th
= 711 + 132
= 843
⇒ Tn = 843
⇒ a+(n-1)d = 843
⇒ 3+(n-1)12 = 843
⇒ (n-1)12 = 843 -3
⇒ (n-1)12 = 840
⇒ (n-1) = 840/12
⇒ (n-1) = 70
⇒ n = 70 +1
∴ n= 71
∴The 71st term is the term that 132 more than the 60th term 
0
2016-02-01T20:43:13+05:30
A=3
d=12

A₆₀=A+59d=711
Let A' be the term 132 more than A₆₀
A+(n-1)d=132 +A₆₀
3+(n-1)12=843
n=71
0