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ghantark
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I would suggest using the formula, for the simple reason that instead of an AP the problem might refer to some fancy formula for an unknown sequence of numbers.

Anyway here we go...

For odd integers till 39, number of integers = 20
Sum = 20^2 = 400
For odd integers till (but not including 13), number of integers = 6
Sum = 6^2 = 36
Difference = 364 = B
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13 & 39 inclusive

sum must be sum upto 39-sum upto 11

11 is (11+1)/2=6th term

39 is 20th term

20^2-11^2=364
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Hi Bunnel,

I am struggling hard to understand why we cant treat this as a function question and take N^2 as a function to calulate 14^2 (14 are total odd integers between 13 to 39 and inclusive). 
Bunuel

ghantark
The Sum of first N consecutive odd integers is N^2. What is the sum of all odd integers between 13 and 39 inclusive.

A. 351
B. 364
C. 410
D. 424
E. 450
The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd integers from 1 to 39, inclusive is \(20^2\);

Since there are 6 odd integers from 1 to 11, inclusive then the sum of all odd integers from 1 to 11, inclusive is \(6^2\);

So, the required sum is \(20^2-6^2=364\).


Answer: B
­
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Hi Bunnel,

I am struggling hard to understand why we cant treat this as a function question and take N^2 as a function to calulate 14^2 (14 are total odd integers between 13 to 39 and inclusive). 
Bunuel

ghantark
The Sum of first N consecutive odd integers is N^2. What is the sum of all odd integers between 13 and 39 inclusive.

A. 351
B. 364
C. 410
D. 424
E. 450
The sum of all odd integers between 13 and 39, inclusive equals to the sum of all odd integers from 1 to 39, inclusive minus the sum of all odd integers from 1 to 11, inclusive.

Since there are 20 odd integers from 1 to 39, inclusive then the sum of all odd integers from 1 to 39, inclusive is \(20^2\);

Since there are 6 odd integers from 1 to 11, inclusive then the sum of all odd integers from 1 to 11, inclusive is \(6^2\);

So, the required sum is \(20^2-6^2=364\).


Answer: B
­
­
Becasue we are told that "The Sum of first N consecutive odd integers is N^2", not that the sum of any N consecutive odd integers is N^2.
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­To determine the sum of all odd integers between 13 and 39, inclusive, we first need to identify these integers. The odd integers in this range are:

13,15,17,19, .... ,39

These integers form an arithmetic sequence where the first term a=13 and the common difference d=2.

We can find the number of terms (n) in this sequence using the formula for the n-th term of an arithmetic sequence: A(n)=a+(n−1)d

Setting A(n)=39, the last term, we can solve for n:

39=13+(n−1)⋅2 => n=14There are 14 odd integers in the range from 13 to 39.

The sum of an arithmetic sequence is given by the formula: S(n)=(n*(a+l))/2 

Substituting the values we found:

S(14)=(14*(13+39))/2 = 364Therefore, the sum of all odd integers between 13 and 39, inclusive, is 364 (B) . 
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Sum of all odd integers between 13 and 39 is an example of an evenly spaced set.

Concept

Sum of evenly spaced set \(=mean*n = \frac{first term + last term}{2}*n\)

\(13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 = n = 14\)

Here, first term = 13, and last term = 39

Sum of evenly spaced set \(= \frac{13+39}{2}*14\\
\\
\frac{52}{2}*14 \)

Answer: [B] 364
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Hi. Why are we not taking 1 to 12 inclusive. Why have we ignored 12 here. Its an even number but why is it being ignored while calculating 6 as the number of terms?.
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Hi. Why are we not taking 1 to 12 inclusive. Why have we ignored 12 here. Its an even number but why is it being ignored while calculating 6 as the number of terms?.

Because we are summing odd integers only, 12 is even, so it is not included.

When counting odd numbers from 1 to 11 inclusive, we get: 1, 3, 5, 7, 9, 11. That gives 6 terms.

12 is not an odd number, so it does not count toward the total.
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Ok got it. So it is like if we are calculating consecutive odd numbers in a set then we should take odd to odd inclusive. And for consecutive even numbers, we should take even to even inclusive.
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I have an easier method to approach this question:
Odd consecutive numbers from 13 to 39 are: 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, and 39 (14 numbers)
Since they are evenly spaces- we can do= Median x No. of terms
Here, Median would be: (25 + 27)/ 2 = 26
and then 26 x 14 (no of terms)= 364

Answer B

Is there any flaw?
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Let's first visually represent 1, 13 and 39:

1-------13----------------39

Now, we are given that the sum of the first n consecutive positive odd integers is equal to n^2.

Note: Here, n refers to the number of the odd terms (and not that particular odd term as such)

2n-1 = 39 (thus, n = 20)
The sum of the first n consecutive positive odd integers till 39 = 400 (n^2 = 20^2 = 400)

Now, if you look at the above number line, we are required to find out the sum of all the consecutive odd numbers between 13 and 39, inclusive. This can also be thought of as the (sum of all consecutive odd numbers from 1 to 39) - (sum of all consecutive odd numbers from 1 to 11).

2n-1 = 11 (thus, here n = 6)
The sum of the first n consecutive positive odd numbers till 11 = 36 (n^2 = 6^2 = 36)

Therefore, the required sum = 400-36 = 364 (Option B) (Answer)
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