NDA 2 2025 Maths, question 33

Topic: Algebra

Passage · Q33 to Q34

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

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

Q33

What is the relation between $m$ and $n$ ?

(a) $n^{2}=m(m+1)$
(b) $n^{2}=m(m+2)$
(c) $n^{2}=2m(m+1)$
(d) $n^{2}=2m(m+2)$ Correct answer

Solution.

Answer: (d) $n^{2}=2m(m+2)$

The first AP has first term 6 and common difference 4.
Its m-term sum is m[12+4(m−1)]/2=2m(m+2).
The sum of n odd numbers is n², giving n²=2m(m+2).