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30 Aug 2022, 16:37
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This is for series and sequences questions
1) Sum of series = Number of terms/2(2(first term)+(number of terms-1)(difference)
2)Last term in sequence= First term+(number of terms-1)(difference)
Highly suggest learning these two in particular as you are able to use them in variety of scenarious.
Example: What's the sum of the even numbers between 99 and 554
Example 2: How many numbers are there between 1 and 12314 (I know this type of question gives people trouble sometimes)
1) Sum of series = Number of terms/2(2(first term)+(number of terms-1)(difference)
2)Last term in sequence= First term+(number of terms-1)(difference)
Highly suggest learning these two in particular as you are able to use them in variety of scenarious.
Example: What's the sum of the even numbers between 99 and 554
Example 2: How many numbers are there between 1 and 12314 (I know this type of question gives people trouble sometimes)
30 Aug 2022, 21:18
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If a is the first term
n is number of terms
l is the last term
d is the common difference
S is sum of first n terms
l = a + (n-1)d
S = n/2 {a + (n-1)d} = n(a+l)/2
n is number of terms
l is the last term
d is the common difference
S is sum of first n terms
l = a + (n-1)d
S = n/2 {a + (n-1)d} = n(a+l)/2
jg3swish
30 Aug 2022, 22:11
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Hi,
To add on to this:
The sum of the series = (2*a + (n-1)*d)*(n/2)
And any term (not just the last) = a + (n-1)*d where n is the position of the term
The solution of the mentioned examples are:
1) What's the sum of the even numbers between 99 and 554:
Required: Sum of 100 (the first even) to 552 (the last even) (inclusive)
Number of terms of the arithmetic progression = (552-100)/spacing(2) + 1
=227 even terms
Sum of 100 to 552 =(2*a + (n-1)*d)*(n/2)
= (2*100+(227-1)*2)(227/2)=74002
We could have arrive at the same number by using Sum of arithmetic progression = mean of series * number of terms
= (100+552)/2*227= 74002
Similar Question:https://gmatclub.com/forum/what-is-the-sum-of-all-the-even-integers-between-99-and-262495.html#p2039611
2)How many numbers are there between 1 and 12314
As Calculated in the above example, for any arithmetic series (evenly distributed, here the numbers are consecutive so the spacing is = 1)
For number of terms from 1 to 12314 (excluding: subtract 1 from diff/spacing)
= (12314-1)/1 -1 =12312
Another approach for number of terms excluding 1 and 12314 i.e number of terms from 2 to 12313 (inclusive: add 1)
= (12313-2)/1 +1 = 12312
Useful Sequence & series thread: https://gmatclub.com/forum/sequences-made-easy-all-in-one-topic-262490.html#p2039594
To add on to this:
The sum of the series = (2*a + (n-1)*d)*(n/2)
And any term (not just the last) = a + (n-1)*d where n is the position of the term
The solution of the mentioned examples are:
1) What's the sum of the even numbers between 99 and 554:
Required: Sum of 100 (the first even) to 552 (the last even) (inclusive)
Number of terms of the arithmetic progression = (552-100)/spacing(2) + 1
=227 even terms
Sum of 100 to 552 =(2*a + (n-1)*d)*(n/2)
= (2*100+(227-1)*2)(227/2)=74002
We could have arrive at the same number by using Sum of arithmetic progression = mean of series * number of terms
= (100+552)/2*227= 74002
Similar Question:https://gmatclub.com/forum/what-is-the-sum-of-all-the-even-integers-between-99-and-262495.html#p2039611
2)How many numbers are there between 1 and 12314
As Calculated in the above example, for any arithmetic series (evenly distributed, here the numbers are consecutive so the spacing is = 1)
For number of terms from 1 to 12314 (excluding: subtract 1 from diff/spacing)
= (12314-1)/1 -1 =12312
Another approach for number of terms excluding 1 and 12314 i.e number of terms from 2 to 12313 (inclusive: add 1)
= (12313-2)/1 +1 = 12312
Useful Sequence & series thread: https://gmatclub.com/forum/sequences-made-easy-all-in-one-topic-262490.html#p2039594
