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In a arithmetic sequence with a^1=12, d=3 which term is equal to 159?

User Blazejmar
by
8.4k points

2 Answers

3 votes

Answer: n=50th term

Explanation:

An= a1+(n-1)d

An= nth term

A1= first term

N= nth position

D= common difference

An =159

A1= 12

D= 3

N= ?

Substitute the values

An=a1+(n-1)d

159=12+(n-1)3

Collect like terms

159-12=(n-1)3

147=3n-3

147+3=3n

150=3n

N=150/3

N= 50

Therefore 50th term will give 156

User Brian Choi
by
8.5k points
1 vote

Answer:

50th term is 159

Explanation:

nth term=159

a + (n -1)*d = 159

12 + (n-1)*3= 159

(n - 1)*3 = 159 - 12

(n - 1)*3 = 147

n -1 = 147/3

n - 1 = 49

n = 49 + 1

n= 50

User Avetis Zakharyan
by
8.2k points

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