CAT 2019 Progression Question (Medium)

Question

If 2n 1 2n 3 2n 5 ... 2n 47 5280 , then what is the value of 1 2 3 ... n ? TITA

Accepted Answer

Step 1 The given sum is 2n 1 2n 3 2n 5 ... 2n 47 This is an arithmetic progression with First term 2n 1 Last term 2n 47 Common difference 2 To find number of terms Last First Common difference 1 2n 47 2n 1 2 1 46 2 1 24 Step 2 Sum of 24 terms 5280 Use sum formula Sum number of terms average of first

CAT 2019

Quantitative

Progression

Medium

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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