Sequence and Series JEE 2024: Why No Finite Minimum Exists

Q: Why can't I cancel a_2023 and a_2024 in the solution?

Consecutive indices do not make two power sums equal. The recurrence gives the exact expression 2(a_2023/a_2024) - s, not 2 - s. For s > 2, the power sums increase strictly, so the retained ratio is less than one.

What does the supplied Sequence and Series JEE 2024 question actually ask?

The supplied Sequence and Series JEE 2024 expression has no finite minimum as written. The worked solution contains an invalid cancellation, so this is a source audit, not a verified PYQ solution.

For a real parameter, take the distinct roots of the quadratic and define their power sums:

x^{2} - (t^{2} - 5 t + 6) x + 1 = 0, t \in R,

α \neq β, a_{n} = α^{n} + β^{n} .

The task is to find the least possible value, wherever defined, of

\frac{a_{2023} - a_{2025}}{a_{2024}} .

The record labels the question 2024 but lists 2023-01-24 Morning; neither the exam date nor shift is verified. No option texts were supplied, so “Correct answer: 1” cannot safely identify either a numerical answer or a mapped option.

The supplied worked solution reports $- \frac{1}{4} .$

Its derivation does not establish that answer.

How do you derive the power-sum recurrence without expanding the powers?

Use the quadratic equation to express each power through the preceding two powers. Both roots obey the same quadratic, so adding their equations gives a recurrence for the power sums. This is the valid part of the supplied solution; no powers near 2024 need to be calculated individually.

First, abbreviate the coefficient. Vieta’s formulas give the sum and product of the roots: $s = t^{2} - 5 t + 6,$

α + β = s, α β = 1 .

Each root satisfies the quadratic separately:

α^{2} = s α - 1, β^{2} = s β - 1 .

For an integer index at least two, multiply by the corresponding root raised to two less than that index:

α^{2} α^{n - 2} = s α α^{n - 2} - α^{n - 2},

β^{2} β^{n - 2} = s β β^{n - 2} - β^{n - 2} .

Adding gives

α^{n} + β^{n} = s (α^{n - 1} + β^{n - 1}) - (α^{n - 2} + β^{n - 2}),

a_{n} = s a_{n - 1} - a_{n - 2} (n \geq 2) .

The starting values follow from the definition and Vieta’s sum. Together, they determine every later term:

a_{0} = 2, a_{1} = s .

Substitute the required index:

a_{2025} = s a_{2024} - a_{2023} .

Provided the denominator is nonzero, $a_{2024} \neq 0,$ the target becomes

\frac{a_{2023} - a_{2025}}{a_{2024}} & = \frac{a_{2023} - (s a_{2024} - a_{2023})}{a_{2024}} & = \frac{2 a_{2023} - s a_{2024}}{a_{2024}} & = 2 \frac{a_{2023}}{a_{2024}} - s .

Keep the neighbouring-term ratio. The recurrence simplifies the expression but does not make those two terms equal.

Does the stated expression have a finite minimum?

No. The expression is unbounded below on an admissible range with distinct positive real roots. That alone rules out a finite minimum as written. We do not need to examine every other parameter value.

Complete the square to see which large coefficient values are available:

s = {(t - \frac{5}{2})}^{2} - \frac{1}{4} .

This coefficient is unbounded above. In particular, every value in the following test range is attainable: $s > 2 .$

Choose the larger root as the first root. The quadratic formula and root product give

α = \frac{s + \sqrt{s^{2} - 4}}{2} > 1, β = α^{- 1} .

These roots are distinct, positive and real. Therefore,

a_{n} = α^{n} + α^{- n}, a_{2024} > 0 .

The target is defined throughout this range. To bound its remaining ratio, subtract consecutive power sums and factor:

a_{n + 1} - a_{n} & = α^{n + 1} + α^{- n - 1} - α^{n} - α^{- n} & = (α - 1) (α^{n} - α^{- n - 1}) & > 0 (n \geq 0) .

Both factors are positive because the larger root exceeds one. Hence the positive power sums increase strictly:

0 < \frac{a_{2023}}{a_{2024}} < 1 .

Substitute this bound into the exact expression:

\frac{a_{2023} - a_{2025}}{a_{2024}} = 2 \frac{a_{2023}}{a_{2024}} - s < 2 - s .

As the coefficient grows, the right-hand side falls below any prescribed real bound. So does the target expression.

The supplied final answer therefore cannot be endorsed without a corrected question or source solution. Guessing a replacement sign, exponent or domain would produce a different problem.

Why can you not cancel neighbouring power sums?

Consecutive indices do not imply equal terms. The first invalid step replaces the retained ratio by one:

2 \frac{a_{2023}}{a_{2024}} - s ≢ 2 - s .

That replacement silently assumes $a_{2023} = a_{2024} .$

The supplied argument never proves this equality. Our positive-root check proves the opposite throughout the test range.

The reciprocal-root condition does not make consecutive power sums equal. Each root generates a geometric sequence, but the sum of two geometric sequences need not have one constant common ratio.

The later optimisation also needs an independent check:

The polynomial opens upward, so its vertex is a minimum, not a maximum.
Negating the expression requires changing every sign inside the brackets.
The vertex value must come from substitution or completing the square.

The correct identities are

f (t) = t^{2} - 5 t - 4 = {(t - \frac{5}{2})}^{2} - \frac{41}{4},

f (\frac{5}{2}) = - \frac{41}{4},

2 - (t^{2} - 5 t + 6) = - (t^{2} - 5 t + 4) .

The cancellation is an error in the provided derivation, not evidence about students’ choices. A wrong-option mapping cannot be supplied because the option list is absent.

Which two original exercises help you check this method?

Practise one numerical recurrence and one expression that genuinely collapses. These original exercises test generating power sums without finding roots and checking signs before cancellation. They are not additional PYQs.

How do you calculate the second and third power sums without finding the roots?

The required values are seven and eighteen. For this first original exercise, calculate the second and third power sums defined by

α, β solve x^{2} - 3 x + 1 = 0, b_{n} = α^{n} + β^{n} .

Vieta’s formulas provide the starting values. The same root-by-root derivation gives

b_{0} = 2, b_{1} = 3,

b_{n} = 3 b_{n - 1} - b_{n - 2} .

Calculate in order:

b_{2} = 3 (3) - 2 = 7,

b_{3} = 3 (7) - 3 = 18 .

Finding the roots adds unnecessary work here. These unequal consecutive terms also show why neighbouring power sums cannot simply be cancelled.

When does a neighbouring-power-sum fraction simplify to the coefficient?

The sum of the neighbouring terms collapses directly through the recurrence. For this second original exercise, use

x^{2} - s x + 1 = 0, b_{n} = α^{n} + β^{n},

and evaluate

\frac{b_{n + 1} + b_{n - 1}}{b_{n}}, n \geq 1, b_{n} \neq 0 .

Rearrange before dividing:

b_{n + 1} = s b_{n} - b_{n - 1},

b_{n + 1} + b_{n - 1} = s b_{n},

\frac{b_{n + 1} + b_{n - 1}}{b_{n}} = s .

The plus sign combines the neighbours into an exact multiple of the denominator. The supplied question’s difference retains a neighbouring-term ratio.

This exercise is not a recovered or corrected paper question. Before cancelling, write the recurrence with all three indices visible, then check the numerator’s signs against it.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

How do I find power sums of quadratic roots without finding the roots?

For roots alpha and beta of x² - sx + 1 = 0, define a_n = alpha^n + beta^n. Each root satisfies the quadratic, giving a_n = s a_{n-1} - a_{n-2} for n ≥ 2, with a_0 = 2 and a_1 = s. This recurrence generates the power sums without calculating either root.

Does the supplied Sequence and Series JEE 2024 expression have a minimum?

No, the expression (a_2023 - a_2025)/a_2024 is unbounded below as written. Set s = t² - 5t + 6; for s > 2, the roots are distinct and positive, and 0 < a_2023/a_2024 < 1. The expression is therefore less than 2 - s, which falls without bound as s increases.

Why can't I cancel a_2023 and a_2024 in the solution?