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Question
Choose the correct alternative answer for the following question .
In an A.P. 1st term is 1 and the last term is 20. The sum of all terms is = 399 then n = ....
Options
42
38
21
19
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Solution
It is given that,
First term (a) = 1
Last term (tn) = 20
Sum of terms (Sn) = 399
We know that,
\[t_n = a + \left( n - 1 \right)d\]
\[ S_n = \frac{n}{2}\left( 2a + \left( n - 1 \right)d \right)\]
\[ \Rightarrow S_n = \frac{n}{2}\left( a + \left( a + \left( n - 1 \right)d \right) \right)\]
\[ \Rightarrow S_n = \frac{n}{2}\left( a + t_n \right)\]
\[ \Rightarrow 399 = \frac{n}{2}\left( 1 + 20 \right)\]
\[ \Rightarrow 399 = \frac{21n}{2}\]
\[ \Rightarrow 21n = 399 \times 2\]
\[ \Rightarrow n = \frac{798}{21}\]
\[ \Rightarrow n = 38\]
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