Question:
The 4th term from the end of an AP – 11, -8,-5,…, 49 is
(a) 37
(b) 40
(c)43
(d) 58
Solution:
(b) We know that, the n th term of an AP from the end is
$a_{n}=l-(n-1) d \quad \ldots$ (i)
Here, $l=$ Last term and $l=49$ [given]
Common difference, $\quad d=-8-(-11)$
$=-8+11=3$
From Eq. (i), $a_{4}=49-(4-1) 3=49-9=40$