Question

Find the 12th term from the end of the following arithmetic progressions:
(i) 3,5,7,9,...201
(ii) 3,8,13,...,253
(iii) 1,4,7,10,...,88

Answer 1

1) First AP = -3,5,7,9,...201
Here, a=201 and d=5-3=2.
According to the question,
d=(-2)[becoz ap stats from the end]
now $T_{12}$ = a +11d =201+11* -2
⇒ a+ 11d =201+(-22)
⇒ a+ 11d = 179


2) Second AP 3,8,13,...,253

a= 3, d= 5
$T_n$ =l=253
⇒ 253=3+(n-1)5
⇒ 250=(n-1)5
⇒ 250=5n-5
⇒ 255=5n
⇒ n=255/5
⇒ n=51

3-(iii) 1,4,7,10,...,88
$T_{12}$ =a+11d
a=1
d=3
$T_{12}$ =1+11*3
$T_{12}$ =33