CAT 2025 Slot 1 · Quant · Question 4

As we know that the sum of first n odd numbers = {n}^{2}

n = \frac{57-1}{2}+1=29 terms, and their total is {29}^{2}=841.

Let k be the m-th term, so k=2m-1.

Everything below k is the first m-1 odd numbers, sum = {\left( m-1 \right)}^{2}.

The two sides are equal, and together with k they make up the whole:

2{\left( m-1 \right)}^{2}+k=841, that is 2{\left( m-1 \right)}^{2}+2m-1=841.

Expanding, 2{m}^{2}-2m-840=0,

so {m}^{2}-m-420=0 and \left( m-21 \right)\left( m+20 \right)=0,

giving m=21.

Hence k=2\left( 21 \right)-1=41.

(Check: below it, {20}^{2}=400; above it, 841-400-41=400.)