Question

2. Find the third last term from the end of the
A.P. -7,0, 7, ..., 161.
​

Answer 1

a=161(Considering from end)

d=0-7=-7

Therefore,

3rd term(from the end) = 161 + (3-1)*(-7)

=161+2*(-7)

=161-14

=147

Answer 2

hey mate

here is your answer

the first AP is -7,0,7,.....,161

let the AP be

161,154,147,.....,-7

a=161, d =-7

a3=161-2*7=161-14=147

hope it helps u