5+9+13+...+49 find the sum ( 324) how do you get the answer?

This is an arithmetic sequence.
First term, a₁ = 5
Common difference, d = 9 - 5 = 13 - 9 = 4
Number of terms = n

an = a₁ + (n - 1)d
49 = 5 + (n - 1) × 4
4(n - 1) = 44
n - 1 = 11
n = 12

5 + 9 + 13 + ...... + 49
= Sum of the 12 terms
= n × (a₁ + an) / 2
= 12 × (5 + 49) / 2
= 324

This is an Arithmetic progression.
Sn = a + (a + d) + ----------[ a + (n - 1) d ]
Sn = [ a + (n - 1) d ] + -------(a + d) + a

2 Sn = n [ 2a + (n - 1) d ]
2 Sn = n [ a + [a + (n - 1) d ] ]
2 Sn = n [ first + last ]
Sn = [ n/2 ] [ first + last ]
S12 = 6 [ 5 + 49 ]
S12 = 6 [ 54 ]
S12 = 324

5, 9, 13, 17, ...
an = 4n + 1
If an = 49, n = 12
Sn = n(a1 + an)/2
5 + 9 + 13 + ... +49
= 12(5 + 49)/2
= 324

you plus all the numbers except 324 together then subtract that number from 324 eg. 324-76