Question

Which term of the ap 8,14,20,26.....wil be 72 more than its 41st term

Answer 1

t1=8
d=t2-t1 = 14-8 = 6
d=6

find tn=?

tn=a+(n-1)xd
t41= 8+(41-1) × 6
t41=8+(40×6)
t41=8+240
t41=248


here 41st term is 248

therfore, 248+72
......... 248+72 = 320


therefore

Answer 2

AP = 8 , 14 , 20 , 26 , ............



Here,


First term ( a ) = 8


Common difference ( d ) = 14-8 = 6


Therefore,


41st term is given by

T41 = a + 40d


T41 = 8 + 40 × 6



T41 = 248.




Required term = 248 + 72 = 320.



Let nth term is 72 more than it's 41st term . Then,




Tn = 248


a + ( n - 1 ) × d = 320


8 + ( n - 1 ) × 6 = 320


8 + 6n - 6 = 320



6n + 2 = 320


6n = 318


n= 53.


Hence,

53th term is the required term.