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AEMCE

This topic has 1 expert reply and 1 member reply
Meek_21 Newbie | Next Rank: 10 Posts
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AEMCE

Post Sun Sep 24, 2017 9:14 pm
Find the sum of all even integers from 12 to 650

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Post Mon Sep 25, 2017 2:49 am
Meek_21 wrote:
Find the sum of all even integers from 12 to 650
For any EVENLY SPACED SET:
Number of terms = (biggest-smallest)/interval + 1, where the interval is the distance between successive terms.
Average = median = (biggest + smallest)/2.
Sum = (number of terms)(average).

Here, the interval between successive terms is 2.
Thus:
Number of terms = (650-12)/2 + 1 = 325 - 6 + 1 = 320.
Average = (650+12)/2 = 325 + 6 = 331.
Sum = 320*331 = 105920.

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pannalal Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Sep 24, 2017 11:49 pm
Find the sum of all even integers from 12 to 650

We can solve question in different ways. Below, I give two methods:

(1) Let us divide the number 12 and 650 by 2 as we want sum of even numbers. We get 6 to 325. We add these numbers and multiply by 2. Total numbers = 325-6+1 = 320. So,
Sum = 320*(6+325)/2*2 = 320*331 = 105920.

(2) Using Arithmetic Progression Sum.
Sum = n/2*{2a+(n-1)d} where n = 320, a = 12, d = 2.
= 320/2*(24 + 319x2) = 160*(24 + 638) = 160*662 = 105920.

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