Question

find the sum of series 1+3+5+.....+399

Answer 1

it is an ap.
so, añ = 1+ (n-1)2
= 399 = 1+2n -2
= 2n = 398
= n = 199
Now, s(199) = 199/2 { 2(1) +198× 2}
solve it...u found the answer

Answer 2

Answer:

Step-by-step explanation:

Let there are “n” term in the series

Therefore $a_n$=a+(n-1)d

Where a =first term and d =common difference

399=1+(n-1)2

399=2n-1

2n=400

n=200

So therefore we can say that

$S_n$=n/2(2a+nd)

Therefore sum of series=40200

hence this is your answer