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CAT 2019 Slot 2 — QA Question 29

Arithmetic ProgressionEasy

If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1 + 2 + 3 + ... + n?

Answer & solution

Correct answer: 4851

Answer: 4851

Solution

Easy

Split the left side into the repeated 2n2n part plus the sum of consecutive odd numbers. Count the terms, use "sum of first mm odd numbers =m2=m^2", solve for nn, then apply the formula for 1+2+⋯+n1+2+\dots+n.

1

Count the terms. The odd parts run 1,3,5,…,471,3,5,\dots,47.

number of terms=47+12=24\begin{aligned} &\text{number of terms} = \frac{47+1}{2} = 24 \end{aligned}
2

Separate the two pieces. Each of the 2424 terms carries a 2n2n; the odd numbers sum to 24224^2 (sum of first 2424 odd numbers).

(2n)⋅24+(1+3+⋯+47)=5280⇒ 48n+242=5280(1+3+⋯+47=242)⇒ 48n+576=5280\begin{aligned} &(2n)\cdot 24 + (1+3+\dots+47) = 5280\\ &\Rightarrow\ 48n + 24^2 = 5280 \quad\text{(}1+3+\dots+47=24^2\text{)}\\ &\Rightarrow\ 48n + 576 = 5280 \end{aligned}
3

Solve for nn.

48n=5280−576=4704⇒ n=98\begin{aligned} &48n = 5280 - 576 = 4704\\ &\Rightarrow\ n = 98 \end{aligned}
4

Compute 1+2+⋯+n1+2+\dots+n.

1+2+⋯+98=98×992⇒ =4851\begin{aligned} &1+2+\dots+98 = \frac{98\times 99}{2}\\ &\Rightarrow\ = 4851 \end{aligned}
1+2+⋯+n=48511+2+\dots+n = 4851

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CAT 2019 Slot 2 QA Q29: If (2n+1) + (2n+3) + (2n+5) + ... + (2n+47) = 5280 , then what is the value of 1 + 2 + 3 + ... + n? — Solution | TheCATExam