SOLUTION: The sum of the first 20 terms of an arithmetic sequence is 840. If the common difference is 4, find (a) The first term (b) The sum of the first 30 terms

Question 819797: The sum of the first 20 terms of an arithmetic sequence is 840. If the common difference is 4, find
(a) The first term
(b) The sum of the first 30 terms
Answer by Ms. U(2) (Show Source): You can put this solution on YOUR website!
Formula: S= n/2(2a+(n-1)d)
where
s = sum of progression
n = number of terms
a = value of first term
d = difference between 2 terms or the common difference
(a)
Given:
d= 4
S(sum of the 20 terms)=840
n= 20 (because there are 20 terms)
a= ???
Substitute the given value.
840 = 20/2[2a+(20-1)4]
840 = 10[2a+ (19)4]
840 = 10 (2a + 76) *Distribute 10 in(2a + 76)*
840 = 20a + 760 *Transpose 760 to the left side of the equal sign.
840 - 760 = 20a
80 = 20a * Divide both sides by 20.
80/20 = 20a/20
a= 4
The first term is 4.

(b)
a= 4
d= 4
n= 30
S(sum of the 30 terms)=???
S= 30/2 [2(4)+ (30-1)4]
S= 15 [8 + (29)4]
S= 15 [8 + 116]
S= 15 (124)
S= 1860
The sum of the first 30 terms is 1860.
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