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b14kumar
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A man has 50 mangoes and he places the 50 mangoes like this: 30 Jun 2006, 10:16
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A man has 50 mangoes and he places the 50 mangoes like this: the distance between 1and 2 is 1 yard and the distance between 2 and 3 is 3 yards and the distance between 3 and 4 is 5 yards; like this he increases the distance between mangoes in steps of two. After arranging them like that and if he start picking them up how much yard he has to walk?

Regards,
Brajesh



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A man has 50 mangoes and he places the 50 mangoes like this: 30 Jun 2006, 12:13
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When n=the number of mango the following gives you the distance:

2n-3 --> distance between mangos
If n=2 --> 2(2)-3=1
If n=3 --> 2(3)-3=3
.
.
.
If n=50 --> 2(50)-3=97

So, the sum of the total distance is: 1+3+5+7+9.....+97=2401

But can someone plz show me the shortcut on how to sum consecutive even or odd numbers??

Thanks.
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A man has 50 mangoes and he places the 50 mangoes like this: 30 Jun 2006, 13:26
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This becomes as A.P.
1,3,5,7........49.

Total terms = n = 49
First term = a = 1
distance between terms = d = 2

Formula for sum = (n*(2a+(n-1)*d))/2 = (49(2*1 + 48 * 2))/2 = 49 * 49 = 2401

Another method:
a[n] = a1 + (n-1) * d

First term = a = 1
Last term = l = 1 + 48 * 2= 97

Sum = n(a+l)/2 = 49(1+97)/2 = 49 * 49 = 2401
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A man has 50 mangoes and he places the 50 mangoes like this: 30 Jun 2006, 21:01
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from the pattern we see that we have an odd incremental...

1+3+5+7+...91+93+95+97

If you add the largest and the smallest you get 98. Keep this pattern going... you will notice that you will end up with 24 of 98 sums. But the middle of the sequence will be at mango 25 and the yardage is 49.

Therefore, 24*98+49 = 2401
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A man has 50 mangoes and he places the 50 mangoes like this: 30 Jun 2006, 21:45
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ps_dahiya
This becomes as A.P.
1,3,5,7........49.

Total terms = n = 49
First term = a = 1
distance between terms = d = 2

Formula for sum = (n*(2a+(n-1)*d))/2 = (49(2*1 + 48 * 2))/2 = 49 * 49 = 2401

Another method:
a[n] = a1 + (n-1) * d

First term = a = 1
Last term = l = 1 + 48 * 2= 97

Sum = n(a+l)/2 = 49(1+97)/2 = 49 * 49 = 2401
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Correct. It's a problem related to A.P

Regards,
Brajesh
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necromonger
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A man has 50 mangoes and he places the 50 mangoes like this: 02 Jul 2006, 08:37
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Sum of n consecutive odd numbers is n^2 :)

so 49 odd => 49^2 is sum = 2401. Shortcut!

for n consecutive even numbers with n terms it is n(n+1)


X & Y
When n=the number of mango the following gives you the distance:

2n-3 --> distance between mangos
If n=2 --> 2(2)-3=1
If n=3 --> 2(3)-3=3
.
.
.
If n=50 --> 2(50)-3=97

So, the sum of the total distance is: 1+3+5+7+9.....+97=2401

But can someone plz show me the shortcut on how to sum consecutive even or odd numbers??

Thanks.
Show more



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!