JEE Main 2022 · 24 Jun · Shift 1 · Mathematics Q13

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MathematicsSingle correct+4 / -1Question 13 of 90

If $\{a_i\}_{i=1}^n$ , where $n$ is an even integer, is an arithmetic progression with common difference 1, and $\displaystyle\sum_{i=1}^na_i=192$ , $\displaystyle\sum_{i=1}^{n/2}a_{2i}=120$ , then $n$ is equal to :

104

Answer and worked solution

Official answer: B

Write the terms. $a_i=a+(i-1)$ .

Two sums. $\sum_{i=1}^na_i=na+\dfrac{n(n-1)}2=192$ , and $\sum_{i=1}^{n/2}a_{2i}=\sum(a+2i-1)=\dfrac n2a+\dfrac{n^2}4=120$ .

Subtract. Doubling the second: $na+\dfrac{n^2}2=240$ . Subtracting the first: $\dfrac{n^2}2-\dfrac{n(n-1)}2=\dfrac n2=48$ , so $n=96$ .

(B)

Topics:Arithmetic progression

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JEE Main 2022 · 24 Jun · Shift 1, Mathematics question 13