In an A.P., if 𝑎 = 1, 𝑎_{n =} 20 and S_n = 399 , then the value of n  is:

We are given the first term of the series, the last, i.e., the 𝑛𝑡ℎ term of the series and the sum of the 𝑛 numbers.

The first number, 𝑎 = 1

The last number, 𝑎 = 20

And the sum of 𝑛 numbers, 𝑆 = 399

We know that the 𝑛𝑡ℎ term of the series, 𝑎    = 𝑎 + (𝑛 − 1)𝑑, where 𝑛 is the total terms

in the series and 𝑑 is the difference between the consecutive terms. Putting the given values in the above equation, we get

20 = 1 + (𝑛 − 1)𝑑

⇒ (𝑛 − 1)𝑑 = 20 − 1

⇒ (𝑛 − 1)𝑑 = 19

Now, the sum of the 𝑛 terms of a series is given by the formula

S_n=  $\frac{n}{2}$(2𝑎 + (𝑛 − 1)𝑑

Putting the values of given values and the value of (𝑛 − 1)𝑑 in the above equation, we get

399 =$\frac{n}{2}$ (2 × 1 + 19)

399 =  $\frac{n}{2}$× 21

n = $\frac{399\times 2}{21}$

n = 38

Hence, the correct option is (c) 38.