In an $A . P ., a = 1, a_{n} = 20$ and $S_{n} = 441$ , then find $n$ .

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Solution

$a = 1, a_{n} = 20$

$S_{n} = 441$

$a_{n} = a + (n - 1) d = 20$

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

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$441 = \frac{n}{2} [2 + 19]$ [substituting the value from (1)]

$n = \frac{441 \times 2}{21} = 42$

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