Question

The first term of an A.P. is 1 and the nth term is
20. If Sn = 399, then n=...
(A) 42 (B) 38 (C) 21 (D) 19​

Answer 1

Answer:

B)38

Step-by-step explanation:

t1=1

tn=20

sn=399

Now,

Sn=n/2*(t1+tn)

399=n/2*(1+20)

399=n/2*21

798=21n

n=798/21

n=38

Answer 2

The value of n is 38.

Step-by-step explanation:

Given : The first term of an A.P. is 1 and the nth term is  20. If Sn = 399

To find : The value of n ?

Solution :

The sum of n terms of A.P is given by,

$S_n=\frac{n}{2}[a+l]$

Where, a is the first term a=1

l is the nth term l=20

$S_n$=399 is the sum of n terms

Substitute the value in the formula,

$399=\frac{n}{2}[1+20]$

$798=21n$

$n=\frac{798}{21}$

$n=38$

Therefore, the value of n is 38.

#Learn more

Question 1 Which term of the A.P. 121, 117, 113 … is its first negative term? [Hint: Find n for an < 0]

brainly.in/question/1345129