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Question 15665: find five consecutive odd intergers such that the sum of the first and fifth number is one less than three times the fourth number
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! the general expression for odd integer is 2n+1...since ..n is any integer, 2n is always even . if we add one to an even number we get an odd number
so let the first odd integer be 2n+1
the second point to be noted is that consecutive odd integers differ by 2 .
( like 1,3,5 etc..)..so now
1st. number = 2n+1
2nd.no.=2n+3
3rd.no. = 2n+5
4th.no. = 2n+7
5th. no. =2n+9
sum of 1st and 5th. nos .=2n+1+2n+9=4n+10.......(1)
three times the 4th. no. = 3*(2n+7)=6n+21
one less than above = 6n+21-1 = 6n+20.....(2)
(1) = (2) ....so 4n+10=6n+20
4n-6n=20-10=10
-2n=10
n=-5...so the five nos.are
1st. number = 2n+1 = 2(*-5)+1 = -9
2nd.no.=2n+3................=-7
3rd.no. = 2n+5................=-5
4th.no. = 2n+7................=-3
5th. no. =2n+9................=-1
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