ISS 2016 Statistics Paper-1 Solution: Question 55 (Lagrange Interpolation on Cumulative Sums)

This post continues our ongoing question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers. Today we tackle a question that looks unusual at first glance because you are never given the individual terms of the sequence directly — only grouped sums — and you still have to recover one specific term using Lagrange's interpolation formula.

Quick Summary

• Topic: Numerical Analysis — Lagrange's Interpolation Formula applied to cumulative sums

• Question Reference: ISS 2016 Statistics Paper-1, Question 55

• Correct Answer: Option (d) — 4

The Question

It is given that u1 = 1, u2 + u3 = 6 and u4 + u5 + u6 + u7 + u8 = 30. What is the approximate value of u4 when computed by using Lagrange's interpolation formula?

• (a) 7

• (b) 6

• (c) 5

• (d) 4

Understanding What the Question Is Really Asking

At first this looks like a plain algebra problem, but it is not. We know u1 individually. We know u2 and u3 only as a combined total of 6. We know u4 through u8 only as a combined total of 30. There is no way to pull out u4 alone using ordinary algebra, because five unknowns are tied up in a single equation. The question is pointing us toward a different tool entirely: treat the sequence as if it were generated by some underlying smooth function of its index, and use interpolation on the sums we do have to estimate the piece we don't.

Step 1: Convert the Given Information into Cumulative Sums

Define Sn as the running total of the sequence up to the nth term, that is, Sn = u1 + u2 + ... + un. This is the key move in the whole problem — once we think in terms of cumulative sums, every piece of information given to us turns into a clean data point:

• S1 = u1 = 1

• S3 = u1 + (u2 + u3) = 1 + 6 = 7

• S8 = S3 + (u4 + u5 + u6 + u7 + u8) = 7 + 30 = 37

So we now have three known values of the function S(n): S(1) = 1, S(3) = 7, and S(8) = 37. Notice that if we can estimate S(4), we automatically get u4, because S(4) = S(3) + u4, which rearranges to u4 = S(4) − S(3).

Step 2: Why Lagrange's Formula Is the Right Tool Here

Our three known points (1, 1), (3, 7) and (8, 37) are not evenly spaced — the gaps are 2 and 5. Formulas like Newton's Forward or Backward Difference method assume equally spaced arguments, so they don't directly apply here.Lagrange's interpolation formula is the one formula in the numerical methods toolkit that works for any set of distinct points, spaced however they like, which is exactly why the question specifically asks for it. We treat S(n) as a polynomial curve passing through these three points and read off its value at n = 4.

Step 3: Write Down Lagrange's Formula for Three Points

For three points (x0, y0), (x1, y1) and (x2, y2), Lagrange's interpolation formula gives the fitted value at any x as:

P(x) = y0 × [(x − x1)(x − x2)] / [(x0 − x1)(x0 − x2)] + y1 × [(x − x0)(x − x2)] / [(x1 − x0)(x1 − x2)] + y2 × [(x − x0)(x − x1)] / [(x2 − x0)(x2 − x1)]

Here we set (x0, y0) = (1, 1), (x1, y1) = (3, 7), (x2, y2) = (8, 37), and we want P(4), i.e. the estimate of S(4).

Step 4: Compute the Three Denominators

These only depend on the x-values (1, 3, 8), not on where we're evaluating, so work them out first and keep them aside:

• (x0 − x1)(x0 − x2) = (1 − 3)(1 − 8) = (−2)(−7) = 14

• (x1 − x0)(x1 − x2) = (3 − 1)(3 − 8) = (2)(−5) = −10

• (x2 − x0)(x2 − x1) = (8 − 1)(8 − 3) = (7)(5) = 35

Step 5: Compute the Three Numerators at x = 4

• (x − x1)(x − x2) = (4 − 3)(4 − 8) = (1)(−4) = −4

• (x − x0)(x − x2) = (4 − 1)(4 − 8) = (3)(−4) = −12

• (x − x0)(x − x1) = (4 − 1)(4 − 3) = (3)(1) = 3

Step 6: Substitute Everything and Simplify

P(4) = 1 × (−4/14) + 7 × (−12/−10) + 37 × (3/35)

Simplify each term one at a time:

• 1 × (−4/14) = −2/7

• 7 × (−12/−10) = 7 × (6/5) = 42/5 (a negative divided by a negative is positive, so this term stays positive)

• 37 × (3/35) = 111/35

Now add −2/7 + 42/5 + 111/35 using a common denominator of 35:

• −2/7 = −10/35

• 42/5 = 294/35

• 111/35 stays as it is

Adding the numerators: (−10 + 294 + 111)/35 = 395/35 = 79/7 ≈ 11.2857

Pro Tip: The step students most often fumble here is the sign on the middle term, 7 × (−12/−10). Both the numerator and denominator are negative, so they cancel to a positive 6/5 — miss that and your final answer flips to something that doesn't match any option. The way this calculation is usually drilled in a classroom setting is to isolate and simplify each Lagrange term completely on its own, with its sign checked explicitly, before combining anything; students who build that habit stop losing marks to sign errors in interpolation questions.

Step 7: Recover u4 From the Interpolated Value

Remember, we interpolated S(4), the cumulative sum up to the 4th term, not u4 itself. To get u4, subtract the already-known S(3):

u4 = S(4) − S(3) = 79/7 − 7 = 79/7 − 49/7 = 30/7 ≈ 4.2857

Since the question asks for the approximate value, we round 4.2857 to the nearest whole number, giving u4 ≈ 4.

Final Answer: Option (d) — 4.This matches the official answer key for Series A of the ISS 2016 Statistics Paper-1, which also lists option (d) as correct for Question 55.

Why This Question Matters

This question tests something beyond rote formula application — it checks whether a candidate recognizes when raw data needs to be reshaped before a standard numerical method can be applied. Converting individual-term information into cumulative sums is a trick that resurfaces across numerical analysis, actuarial statistics, and time-series smoothing, wherever you're handed partial or grouped totals instead of clean individual observations. Spotting this transformation quickly, rather than getting stuck trying to solve five unknowns from one equation, is exactly the kind of pattern recognition that separates a fast, confident attempt from a stalled one in the exam hall.

Frequently Asked Questions

Why can't we just use simultaneous equations to find u4 directly?

Because the given information provides only two equations (u2 + u3 = 6 and u4 + u5 + u6 + u7 + u8 = 30) for seven unknowns in total. Simultaneous equations need as many independent equations as unknowns, so algebra alone cannot isolate u4. That is precisely why the question directs you toward an interpolation-based approach instead.

Why do we use cumulative sums instead of working with the u-values directly?

The individual u-values are not known as clean data points — only grouped totals are available. By defining S(n) as the running total, every piece of given information (a single value or a group sum) becomes a usable coordinate (n, S(n)), which lets us interpolate a smooth curve through them and then recover any individual term by subtracting consecutive cumulative sums.

Could we have used Newton's Forward Difference Formula instead of Lagrange's?

No, not directly. Newton's forward and backward difference formulas require equally spaced arguments, but our three points correspond to n = 1, 3 and 8, which are unevenly spaced. Lagrange's formula is specifically designed to work with any set of distinct points regardless of spacing, which is why it is the appropriate tool for this data.

Is the answer u4 = 30/7 treated as exact, or is rounding expected?

Rounding is expected. The question explicitly asks for the "approximate value," which is a strong hint that the interpolated result will not come out as a clean integer. Here 30/7 works out to approximately 4.29, and rounding to the nearest whole number gives 4, which is option (d).

What would change if we were asked to find u5 or u6 instead of u4?

The method stays identical, but the target point for evaluating P(x) would shift. For u5, you would compute S(5) using the same three known points and then subtract the interpolated S(4) (not the exact S(3)) to get u5, since S(4) itself would now be an estimate rather than a given value. This compounding of estimates is worth keeping in mind if a question asks for a term further from the known data points.

Does the degree of the interpolating polynomial matter here?

Yes. With exactly three known points, Lagrange's formula fits a quadratic (degree 2) polynomial through them, which is the implicit assumption behind this calculation. If a fourth data point were available, the fitted curve would become cubic and could shift the estimate for u4 slightly, so the "approximate" nature of the answer reflects this polynomial assumption as much as the rounding.

If any step here felt shaky, re-read Step 1 slowly — the entire question hinges on correctly converting the given totals into S(1), S(3) and S(8). Drop your doubts in the comments, and if this walkthrough helped, share it with a fellow ISS aspirant working through the same paper.