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Consecutive odd integers whose sum is 25...

Consecutive odd integers whose sum is `25^2-11^2` are

A

`n=14`

B

`n=16`

C

first odd number is 23

D

last odd number is 49

Text Solution

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The correct Answer is:
A, C, D
Let n consecutive odd numbers be `2k+1,2k+3,2k+5,"....."2k+2n-1`
According to question, sum of these n numbers
`=(n)/(2)[2k+1+2k+2n-1]=n(2k+n)`
`=n^(2)+2kn=(n+k)^(2)-k^(2)`
Given that, `(n+k)^(2)-k^(2)=25^(2)-11^(2)`
`implies n+k=25` and `k=11 " "impliesmn=14` and `k=11` So, first term `=2k+1=23`
Last term `=2k+2n-1=22+28-1=22+27=49`
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