A customer pays ₹975 in instalments. The payment is done each month ₹5 less than the previous month. If the first instalment is ₹100, how much time will be taken to pay the entire amount?

Correct — D. The instalments fall by a fixed ₹5, so they are an arithmetic progression with a = ₹100 and d = −5.

Sₙ = n⁄2 [2a + (n − 1)d]
975 = n⁄2 [200 − 5(n − 1)]
1,950 = 205n − 5n²
n² − 41n + 390 = 0

(n − 15)(n − 26) = 0, so n = 15 or n = 26 — two positive whole roots, and the paper prints both.

Only 15 keeps every instalment a real payment: ₹100, ₹95, ₹90 … ₹30, and 15 × (100 + 30)⁄2 = ₹975 → option (d).

Why the others are wrong

• (a)
  26 months — 26 is the second root of the same quadratic, so algebra alone will not reject it. The schedule does: the 21st instalment is 100 − 5 × 20 = ₹0 and months 22 to 26 are negative, which is money handed back, not paid.
• (b)
  14 months — Fourteen instalments total 14⁄2 [200 + 13(−5)] = 7 × 135 = ₹945, which is ₹30 short. That ₹30 is exactly the 15th instalment.
• (c)
  27 months — Twenty-seven instalments come to ₹945, not ₹975. Sₙ is a quadratic symmetric about n = 20.5, and 14 + 27 = 41, so S₂₇ = S₁₄. By then the payments are long since negative.

A payment falling by a fixed amount each month is an arithmetic progression: here a = 100 and d = −5, so the nth instalment is ₹(105 − 5n).

The sum Sₙ = n⁄2 [2a + (n − 1)d] is a quadratic in n. A target total therefore has two roots, and when d is negative both can be positive whole numbers — S₁₅ and S₂₆ are both ₹975.

The algebra cannot choose between them. The situation does: an instalment is money paid, so a schedule cannot run past its last positive term, the ₹5 paid in month 20.

Key facts

• The instalments form an AP with a = ₹100 and d = −₹5, so the nth instalment is ₹(105 − 5n).
• Sₙ = n⁄2 [2a + (n − 1)d], which here reduces to Sₙ = (205n − 5n²)⁄2.
• n² − 41n + 390 = 0 factorises as (n − 15)(n − 26), so S₁₅ = S₂₆ = ₹975.
• The 21st instalment works out at ₹0 and every later one is negative, so a paying schedule cannot run past 20 months.

Common traps

• Solving the quadratic, spotting 26 among the options and stopping there.
• Reading "₹5 less than the previous month" as ₹5 less than the first month, which makes every instalment after the first ₹95.
• Writing Sₙ = n⁄2 [2a + nd] instead of (n − 1)d. The equation becomes n² − 40n + 390 = 0, whose roots, 20 ± √10, are not whole numbers.

No directly related past PYQ was found.