-  a = 4 ,    l = 199 ,    d = 5
Than, number of terms = 40
Sum of 40 numbers 4060.

We have to find the sum of the series: 4+9+14+...+199

To solve this problem, we can use the formula for the sum of an arithmetic progression. An arithmetic progression is a sequence of numbers in which each term is obtained by adding a constant to the preceding term.

Formula to find the sum of an arithmetic progression:

S = n/2[2a+(n-1)d]
Where,
S = Sum of the series
a = First term of the series
d = Common difference between two consecutive terms
n = Number of terms

In this case,
a = 4
d = 5
n = 40

Therefore,
S = 40/2[2*4+(40-1)5]
= 40/2[8+195]
= 40/2[203]
= 40*203/2
= 4060

Hence, the sum of the series 4+9+14+...+199 is 4060.

If you think the solution is wrong then please provide your own solution below in the comments section .