Question 506885: consider an arithmetc sequence whose fourth entry is 2 and whose seventh entry is 4. find the first and second entries of the sequence.
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! 4th term = 2
7th term = 4
First term = ?
Second term = ?
Use
Tn=a+(n-1)d
Where
Tn = Term
n = Term number
a = First term
d = Common Difference
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For 4th term
a=?, d=?, Tn=2, n=4
Tn=a+(n-1)d
Put the values in above formula
2=a+(4-1)d
2=a+(3)d
2=a+3d ................(1)
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For 7th term
a=?, d=?, Tn=4, n=7
Tn=a+(n-1)d
Put the values in above formula
4=a+(7-1)d
4=a+(6)d
4=a+6d...............(2)
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Subtract (1) from (2)
4=a+6d...............(2)
-2=-a-3d ................(1)
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2=3d
2/3=d
Put the value of d in (1)
2=a+3d ................(1)
2=a+3*(2/3)
2=a+2
2-2=a
a=0
First term = 0
Now
For 2nd Term
Tn=a+(n-1)d
Tn=0+(2-1)2/3
Tn=2/3
For 3rd Term
Tn=0+(3-1)2/3
Tn=0+(3-1)2/3
Tn=4/3
For 4th Term
Tn=0+(4-1)2/3
Tn=0+(3)2/3
Tn=2
and so on ............
Required A.P
0,2/3,4/3, 2,.............
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