Q.A Recurring Deposit of ₹100 per month at 8% per annum matures to ₹2,600. Find the number of monthly instalments.

Since $n$ appears in the formula both linearly and inside $n(n+1)$, substituting the given values reduces the problem to a quadratic equation in $n$, solved exactly as in the Class 11 Quadratic Equations chapter.

$2{,}600 = 100n + \dfrac{100\times8\times n(n+1)}{2400} = 100n + \dfrac{n(n+1)}{3}$ …

Here $P=₹100$, $r=8\%$, $M=₹2{,}600$, and the tenure $n$ (in months) is unknown.

Step 1 — Substitute into the RD formula.

Step 2 — Clear the fraction. Multiplying every term by 3:

Step 3 — Form the quadratic equation.

Step 4 — Solve, exactly as in the Class 11 Quadratic Equations chapter. This factorises as

$(n - 24)(n + 325) = 0 \;\Rightarrow\; n = 24 \text{ or } n = -325$ …

An alternative to factorising by inspection is applying the quadratic formula directly: $n = \dfrac{-301+\sqrt{301^2+4\times7{,}800}}{2} = \dfrac{-301+\sqrt{121{,}801}}{2} = \dfrac{-301+349}{2} = 24$, taking only the positive …

Students sometimes keep both roots of the quadratic, or report the negative root as "the tenure is $-325$ months," without recognising that a tenure must be a positive quantity — the negative root is always mathematically valid fo …