Q44
2 marksSection E

  1. (iii) (a) What is the total distance the competitor has to run ?

Arithmetic Progressions
Sum of first n terms of an AP
Official Answer

Total distance = 370 m. The to-and-fro distances for the 10 potatoes form an AP 10, 16, 22, … with a=10,d=6,n=10a = 10, d = 6, n = 10. Using Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d], S10=5[20+54]=5×74=370 mS_{10} = 5[20 + 54] = 5 \times 74 = 370\text{ m}.

370 msum of APS_n = n/2[2a+(n-1)d]a = 10d = 6n = 10total distance10 potatoes

Marking Scheme

  • 11 mark: correct identification of the AP with a=10,d=6a = 10, d = 6 and n=10n = 10 (or listing the terms up to the 10th).
  • 21 mark: correct substitution and evaluation of S10=102[2(10)+9(6)]=370 mS_{10} = \frac{10}{2}[2(10) + 9(6)] = 370\text{ m}.
  • 3Accept full marks if the student adds the ten terms 10+16+…+6410 + 16 + \ldots + 64 directly to get 370 m.

Hint

Each potato's to-and-fro distance forms the AP 10, 16, 22, … with a=10,d=6,n=10a = 10, d = 6, n = 10; use Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].

Quick Oral Answer

The to-and-fro distances form the AP 10, 16, 22, … with a=10,d=6a = 10, d = 6 and n=10n = 10, so the total distance is S10=102[2(10)+9(6)]=370 mS_{10} = \frac{10}{2}[2(10) + 9(6)] = 370\text{ m}.

Analysis & Explanation

This is the core of the case study — summing an entire AP, not finding a single term.


Concept

  • The to-and-fro distances 10, 16, 22, … form an AP with a=10,d=6,a = 10, d = 6, and n=10n = 10 (ten potatoes in the line).
  • Total distance = Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d], giving S10=5[20+54]=370 mS_{10} = 5[20 + 54] = 370\text{ m}.

Key points

  • Reading n = 10 correctly from "ten potatoes in the line" is essential before applying the formula.

Common mistakes (परीक्षा में सावधानी)

  • Confusing the nth-term formula with the sum formula: computing a_10 = a + 9d = 64 m only gives the last trip, not the total distance.

Real-world

  • This cumulative-distance modelling mirrors real settings like fuel used across repeated delivery runs or total wire cut in equal-step patterns.

Common Mistakes

  1. 1Using the nth-term formula a10=a+9d=64 ma_{10} = a + 9d = 64\text{ m} and reporting 64 m instead of summing the whole AP.
  2. 2Taking n=9n = 9 or n=11n = 11 by miscounting the ten potatoes.
  3. 3Forgetting the factor of 2 in the base terms, i.e., using a=5a = 5 and d=3d = 3 (one-way distances) and getting half the correct sum.

Interesting Facts

The sum 370 m equals ten times the average trip (37 m), illustrating the shortcut Sn=n×(first+last)2=10×(10+64)2S_n = n \times \frac{(\text{first} + \text{last})}{2} = 10 \times \frac{(10 + 64)}{2}.

This exact potato-race sum is a textbook application of Gauss's pairing trick for arithmetic series.

If even one more potato were added, the total would jump by 2×(5+30)=70 m2 \times (5 + 30) = 70\text{ m}, showing how quickly AP sums grow.

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Frequently Asked Questions

Why is this a sum of an AP and not just one term?

The competitor runs a trip for every potato, so the total distance is the sum of all ten trip-distances 10+16+…+6410 + 16 + \ldots + 64, which is the sum of an AP.

How many potatoes are there?

There are ten potatoes in the line, so n=10n = 10 in the sum formula.

What is the shortcut to check the answer?

Use Sn=n(first term+last term)2=10(10+64)2=10×37=370 mS_n = \frac{n(\text{first term} + \text{last term})}{2} = \frac{10(10 + 64)}{2} = 10 \times 37 = 370\text{ m}, matching the main formula.