By signing in you confirm you are 18 or older, or have your parent’s or guardian’s consent.

NDA 2 2025 Maths, question 34

Topic: Algebra

Passage · Q33 to Q34

Let (6+10+14+… up to m terms)=(1+3+5+7+… up to n terms)(6+10+14+\ldots \text{ up to }m\text{ terms})=(1+3+5+7+\ldots \text{ up to }n\text{ terms})

where m<25m<25 and n<25n<25.

Q34

How many values of mm are possible?

  1. (a) None
  2. (b) One
  3. (c) TwoCorrect answer
  4. (d) More than two

Solution.

Answer: (c) Two

  1. Rewrite the relation as n²+2=2(m+1)²; n must be even, so n=2r with 1≤r≤12.
  2. Testing (m+1)²=2r²+1 for these twelve values gives squares only at r=2 and r=12.
  3. Thus (m,n)=(2,4) and (16,24), both within the limits, and two values of m are possible.