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In a sequence of 10 increasing consecutive integers, the sum

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netcaesar
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In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   14 Nov 2006, 13:31
00:00
A
B
C
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E

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In a sequence of 10 increasing consecutive integers, the sum of the last 5 integers is 230. What is the sum of the first 5 integers in the sequence?

A) 225

B) 220

C) 215

D) 210

E) 205



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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   14 Nov 2006, 13:41
E from me.
let the int be x, x+1, x+2....x+9.

x+5+x+6+x+7+x+8+x+9=230
5x+35=230
x=39

sum of first 5=205
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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   14 Nov 2006, 20:32
Answer E.

Let the first 5 number in this sequence be n, n+1, n+2, n+3, n+4. Sum = 5n +10.

Let the last 5 numbers in the sequence be n+5, n+6, n+7, n+8, n+9. Sum= 5n+35.

Since 5n+35 = 230 and we need to find 5n+10. subtract 25 from 230, we have 205 => E
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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   14 Nov 2006, 22:28
First integer, x
Second integer, X+1
third interger, x+2 and so on...

Sum of last five integers is:
x+5 + x+6 + x+7 + x+8 + x+9 = 5x + 35 = 230 --> x = 39

so sum of first five integers is:
x + x+1 + x+2 + x+3 + x+4 = 5x + 10 = 205

ans E
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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   14 Nov 2006, 23:19
Divide 230 by 5 = 46, this is the eighth term, then count back to find the first five and then add.

A:39
B:40
C:41
D:42
E:43
F:44
G:45
H:46
I:47
J:48

If you use formulas it might be quicker, but it's just another way.
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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink] New post   16 Jun 2009, 04:11
The quickest way is first to divide 230 by 5 (to get the mean of the 5 largest numbers) and then subtract 5 from that number and multiply by 5.

Answer is E



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Re: In a sequence of 10 increasing consecutive integers, the sum [#permalink]
16 Jun 2009, 04:11
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