Question

In an $A.P.,a=1,a_n=20\text{and}{s}_n=441,$ then find n.

Answer 1

Given $a=1,a_n=20,3_n=441$

$a_n=a(n-1)d$

$20=a+(n-1)d$

$20=1+(n-1)d$

$\Rightarrow 19=(n-1)d$

Now $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$441=\frac{n}{2}[2\times 1+19]$

$2\times 441=n\left[21\right]$

$\Rightarrow n=42$