SOLUTION: In an arithmetic progression, the sum of the first ten terms is 50 and the 5th term is three times the second term. Find the A. First term. B. Common difference. C. The sum of the

Algebra ->  Sequences-and-series -> SOLUTION: In an arithmetic progression, the sum of the first ten terms is 50 and the 5th term is three times the second term. Find the A. First term. B. Common difference. C. The sum of the       Log On


   

Question 1162188: In an arithmetic progression, the sum of the first ten terms is 50 and the 5th term is three times the second term. Find the A. First term. B. Common difference. C. The sum of the first 100 terms
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

an arithmetic progression:
a%5B1%5D, a%5B1%5D%2Bd, a%5B1%5D%2B2d,.... where a%5B1%5D is the first term and d is Common difference

nth term formula:

a%5Bn%5D=a%5B1%5D%2Bd%28n-1%29

Denote this partial sum by S%5Bn%5D . Then
S%5Bn%5D=n%28a%5B1%5D%2Ba%5Bn%5D%29%2F2 ,where n is the number of terms, a%5B1%5D is the first term and a%5Bn%5D is the last term

given: the sum of the first ten terms is 50+
-> n=10
50=10%28a%5B1%5D%2Ba%5B10%5D%29%2F2
50=5%28a%5B1%5D%2Ba%5B10%5D%29
50=5%28a%5B1%5D%2Ba%5B10%5D%29
10=a%5B1%5D%E2%80%89%2B%E2%80%89a%5B10%5D.......solve for a%5B1%5D
a%5B1%5D=10-a%5B10%5D%E2%80%89....a%5B10%5D%E2%80%89=a%5B1%5D%2B9d
a%5B1%5D=10-%28a%5B1%5D%2B9d%29
a%5B1%5D=10-a%5B1%5D-9d
a%5B1%5D%2Ba%5B1%5D=10-9d
2a%5B1%5D=10-9d
a%5B1%5D=%2810-9d%29%2F2.....eq.1


the 5th term is three times the second term:
a%5B5%5D=3%2Aa%5B2%5D->a%5B2%5D=a%5B1%5D%2Bd and a%5B5%5D=a%5B1%5D%2B4d

a%5B1%5D%2B4d=3%28a%5B1%5D%2Bd%29
a%5B1%5D%2B4d=3a%5B1%5D%2B3d
4d-3d=3a%5B1%5D-a%5B1%5D
d=2a%5B1%5D......... ..eq.2

substitute in eq1
a%5B1%5D=%2810-9%282a%5B1%5D%29%29%2F2.....eq.1
a%5B1%5D=5-9%282a%5B1%5D%29%2F2
a%5B1%5D=5-9a%5B1%5D
a%5B1%5D%2B9a%5B1%5D=5
10a%5B1%5D=5
a%5B1%5D=5%2F10
highlight%28a%5B1%5D=1%2F2%29 -> First term
go to eq.2
d=2a%5B1%5D......... ..eq.2
d=2%281%2F2%29
highlight%28d=1%29->Common difference

nth term formula:
a%5Bn%5D=1%2F2%2B1%28n-1%29
first 10 terms are:
1%2F2, 1%2F2%2B1, 1%2F2%2B2, 1%2F2%2B3,1%2F2%2B4,1%2F2%2B5,1%2F2%2B6, 1%2F2%2B7,1%2F2%2B8,1%2F2%2B9
or
0.5, 1.5, 2.5, 3.5,4.5,5.5,6.5, 7.5,8.5,9.5
check their sum:
S%5B10%5D=10%280.5%2B9.5%29%2F2
S%5B10%5D=5%2810%29
S%5B10%5D=50-> true

The sum of the first 100 terms:
first find 100th term:
a%5B1%5D%2B99d-> 0.5%2B99%2A1=99.5
then the sum is:
S%5B100%5D=100%280.5%2B99.5%29%2F2
S%5B100%5D=50%28100%29
S%5B100%5D=5000


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

In an arithmetic progression, the sum of the first ten terms is 50 and the 5th term is three times the second term. Find the A. First term. B. Common difference. C. The sum of the first 100 terms
Sum of an A.P.: matrix%281%2C3%2C+S%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29

                 ------ eq (i)



Term of an A.P.: matrix%281%2C3%2C+a%5Bn%5D%2C+%22=%22%2C+a%5B1%5D+%2B+%28n+-+1%29d%29

                 

Since matrix%281%2C3%2C+a%5B5%5D%2C+%22=%22%2C+3%28a%5B2%5D%29%29, we get: 



10 = d + 9d ---- Substituting matrix%281%2C3%2C+d%2C+for%2C+2a%5B1%5D%29 in eq (i)

10 = 10d







Sum of an A.P.: matrix%281%2C3%2C+S%5Bn%5D%2C+%22=%22%2C+%28n%2F2%29%282a%5B1%5D+%2B+%28n+-+1%29d%29%29