MathsCBSE Class 10 · Standard · Board exam 2024, set 30/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2024: question 37

Set in the CBSE Class 10 board exam of 2024, paper 30/4/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedArithmetic Progressions

Treasure Hunt is an exciting and adventurous game where participants follow a series of clues/numbers/maps to discover hidden treasures. Players engage in a thrilling quest, solving puzzles and riddles to unveil the location of the coveted prize.

While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forming an A.P. If the number on the $n^{t h}$ spot is $20 + 4 n$ , then answer the following questions to help the players in spotting the clues :

(i)
Which number is on first spot ?
1 mark
(ii)
(a) Which spot is numbered as 112 ?
2 marks
OR
(ii)
(b) What is the sum of all the numbers on the first 10 spots ?
2 marks
(iii)
Which number is on the $(n -2)^{t h}$ spot ?
1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/4/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The rule $20 + 4 n$ gives the number on any spot; just substitute the spot number.

Hint 2

To find which spot carries 112, set $20 + 4 n = 112$ and solve for n.

Hint 3

For the sum, find a and d, then use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . For the last part, replace n by $(n - 2)$ in the rule.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) $a_{n} = 20 + 4 n$ , so $a_{1} = 20 + 4 \times 1 = 24$
1 mark
(ii)(a) Put $a_{n} = 112$ : $20 + 4 n = 112$
1 mark
(ii)(a) $4 n = 92 \Rightarrow n = 23$ , so the 23rd spot is numbered 112.
1 mark
(ii)(b) $a_{2} = 28$ , so $d = a_{2} - a_{1} = 28 - 24 = 4$
½ mark
(ii)(b) $S_{10} = \frac{10}{2} [2 \times 24 + (10 - 1) \times 4] = 5 (48 + 36)$
1 mark
(ii)(b) $S_{10} = 420$
½ mark
(iii) $a_{n - 2} = 20 + 4 (n - 2) = 4 n + 12$
1 mark

Answer: (i) 24; (ii)(a) 23rd spot; (ii)(b) 420; (iii) $4 n + 12$

Where the marks are

(i) First spot number 24 (1 mark)
(ii)(a) Sets up $20 + 4 n = 112$ (1 mark)
(ii)(a) Gets n = 23 (1 mark)
(ii)(b) Finds d = 4 (½ mark)
(ii)(b) Correct use of $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ (1 mark)
(ii)(b) Gets $S_{10} = 420$ (½ mark)
(iii) $(n - 2)$ th term $= 4 n + 12$ (1 mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Finding nth Term from the Sum of n Terms

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/4/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2024: question 37

Hints

Step-by-step solution

Revise the chapter

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