Help Please

[dreamermyrescue]

If a1= 4 an=199 sn=4060 what would the common difference be?
i need work to be shown

[JJ109]

If a1 = 4
a2 = 4 + d (where "d" is the common difference)
a3 = 4 + d + d = 4 + 2d
and so on
an therefore = 4 + (n-1)d
Therefore 4 + (n-1)d = 199
(n-1)d = 199 - 4 = 195
(n-1)d = 195
Call this Equation 1

Then we also know from the standard sum equation of an arithmetical sequence that
sn = (a1 +an) * (n/2) ie the first and last terms of the sequence multiplied by half the number of terms
(4 + 199)* (n/2) = 4060
203 * (n/2) = 4060
n/2 = 4060/203 = 20
n = 40
But we know from Equation 1 that (n-1)d = 195
Therefore (40 -1) d = 195
39d = 195
d = 5

The Common Difference is 5

[Canary42]

Go to the top of the class jj ;-)