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konstantin123 [22]
3 years ago
5

An arithmetic sequence has first term a = 5 and common difference d = 4. how many terms of this sequence must be added to get 41

85?
Mathematics
2 answers:
Zielflug [23.3K]3 years ago
5 0
General Equation : an = a1 + d(n - 1)

Find the nth sequence: 
<u>Given that a1 = 5 and d = 4,</u>
an = 5 + 4(n - 1)
an = 5 + 4n - 4
an = 4n + 1

Find an:
<u>Given that Sn = 4185, find an</u>
sn = (a + an)/2
4185 = (5 + an)/2
5 + an = 8370
an = 8370 - 5
an = 8365

Find n:
<u>We know that an = 8365, find n</u>
an = 4n + 1
8365 = 4n + 1
4n = 8365 - 1
4n = 8364
n = 2091

Answer: 2091 terms
Stels [109]3 years ago
3 0
Given that in arithmetic sequence a=5 and common difference, d=4. Thus the number of terms of this sequence that must be added to attain 4185 will be found as follows:
the explicit formula for arithmetic sequence is given by:
sn=n/2(2a+(n-1)d)
where n is the number of terms:
thus plugging in the values we get:
4185=n/2(2*5+(n-1)4)
solving for n we get:
8370=(10+4n-4)
8370=(6+4n)
4n=8364
n=8364/4
n=2091
thus the number of terms will be:
2091
  
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