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Question: If the number of consecutive odd integers whose sum can be expressed as \({50^2} - {13^2}\) is \(k\)...

If the number of consecutive odd integers whose sum can be expressed as 502−132{50^2} - {13^2} is kk,then kk can be
A.{\text{A}}{\text{.}} 33
B.{\text{B}}{\text{.}} 35
C.{\text{C}}{\text{.}} 37
D.{\text{D}}{\text{.}} 39

Explanation

Solution

Hint: To calculate k value use the arithmetic formula an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d where a is first term, d is common difference, n is number of term and an{{\text{a}}_n} is last term.

Complete step-by-step answer:

It is given in question that kk number of consecutive odd integer is 502−132{50^2} - {13^2}
Therefore,
First series is taken as sum of odd nn consecutive numbers =n2 = {n^2}
Second series is taken as sum of odd mm consecutive numbers =m2 = {m^2}

Therefore, [1+3+5+.....+(2n−1)]−[1+3+5+.....(2m−1)]\left[ {1 + 3 + 5 + ..... + \left( {2n - 1} \right)} \right] - \left[ {1 + 3 + 5 + .....\left( {2m - 1} \right)} \right]=n2−m2{n^2} - {m^2}

It is given in question that,
n2−m2=502−132{n^2} - {m^2} = {50^2} - {13^2}
502−132=(1+3+5...99)−(1+3+5...25)=k k=(27+29...99)  {50^2} - {13^2} = \left( {1 + 3 + 5...99} \right) - \left( {1 + 3 + 5...25} \right) = k \\\ k = \left( {27 + 29...99} \right) \\\
Now using the formula an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
99=27+(n−1)2 72=(n−1)2 36=n−1 37=n n=k  99 = 27 + \left( {n - 1} \right)2 \\\ 72 = \left( {n - 1} \right)2 \\\ 36 = n - 1 \\\ 37 = n \\\ n = k \\\

Counting from 27 till 99 makes k=37k = 37 terms

So, option C is correct answer

Note: In such a type of question the concept of arithmetic progression is used, Consecutive odd integers are arithmetic progression with the difference between each two consecutive terms is two and the first term is odd integer.