The 4th term from the end


The 4th term from the end of an AP – 11, -8,-5,…, 49 is

(a) 37                

(b) 40                    


(d) 58


(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, $\quad d=-8-(-11)$


From Eq. (i), $a_{4}=49-(4-1) 3=49-9=40$


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