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If an A.P. has a = 1, tn = 20 and sn = 399, then value of n is :
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {S_n} = \frac{1}{2}\left( {a + l} \right) \times n \cr
& \Rightarrow 399 = \left( {1 + 20} \right) \times \frac{n}{2} \cr
& \Rightarrow 399 \times 2 = 21 \times n \cr
& \Rightarrow n = 399 \times \frac{2}{{21}} \cr
& \Rightarrow n = 19 \times 2 \cr
& \Rightarrow n = 38 \cr} $$
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