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Q.
Determine the number of terms in the $A.P. 3, 7, 11,..., 407$. Also, find its $20^{th}$ term from the end.
Sequences and Series
Solution:
Here, first term $a = 3$ and common difference $d = 4$.
Let there be $n$ terms in the given $A.P$. Then,
$ n ^ {th} $ term $= 407 $
$\Rightarrow 407 = 3 + (n - 1) \times 4 $
$\Rightarrow n= 102$
So, there are $102$ terms in the given $A.P$.
Now, $20^{th}$ term from the end
$= [102 - 20 + 1]^{th}$ term from the beginning
$= 83^{rd}$ term from the beginning
$ = 3 + (83 - 1) \times 4 = 331$