# CAT 2019 Question Paper | Quants Slot 2

###### CAT Previous Year Paper | CAT Quants Questions | Question 29

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Question 29 : If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1+2+3+ ... +n ? [TITA]

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Given series - (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280
Isolate 2n terms on one side
(2n + 2n +.... + 2n) + (1 +3 + 5 +.... + 47) = 5280
Odd numbers from 1 to 47 are added in the above series.
Number of terms from 1 to 47 = 24 terms
Therefore, the number of 2n terms = 24
For computing the value of (1 +3 + 5 +.... + 47),
We know 1 = 12, 1 +3 = 22, 1 +3 +5 = 32 and so on
So, (1 +3 + 5 +.... + 47) = 242
So, 2n x 24 + 242 = 5280
2n x 24 = 5280 - 242
2n = 220 - 24
n = 110 - 12
n = 98
So, value of 1+2+3+...+98 = = = Value of 1+2+3+...+98 = 4851

The question is "If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1+2+3+ ... +n ? [TITA] "

##### Hence, the answer is 4851

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