MathsCBSE Class 10 · Standard · Board exam 2023, set 30/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 35(a)

Set in the CBSE Class 10 board exam of 2023, paper 30/1/1, section D, question 35(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(a) · 5 marks · Long answerArithmetic Progressions

The ratio of the $1 1^{t h}$ term to $1 7^{t h}$ term of an A.P. is $3 : 4$ . Find the ratio of $5^{t h}$ term to $2 1^{s t}$ term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/1/1.

Hints

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

Hint 1

Write the 11th and 17th terms using $a + (n - 1) d$ and set up the given ratio.

Hint 2

Solving the ratio gives $a$ as a multiple of $d$ . Substitute this into every later expression so that $d$ cancels.

Hint 3

For the sums use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with $n = 5$ and $n = 21$ .

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.

Let the first term be $a$ and the common difference $d$ . Using $a_{n} = a + (n - 1) d$ :
$\frac{a _{11}}{a _{17}} = \frac{a + 10 d}{a + 16 d} = \frac{3}{4}$
1 mark
Cross-multiplying: $4 a + 40 d = 3 a + 48 d$ , so $a = 8 d$ ...(i)
(Here $d \neq = 0$ ; otherwise the ratio would be $1$ , not $\frac{3}{4}$ .)
1 mark
$\frac{a _{5}}{a _{21}} = \frac{a + 4 d}{a + 20 d} = \frac{8 d + 4 d}{8 d + 20 d} = \frac{12 d}{28 d} = \frac{3}{7}$
So $a_{5} : a_{21} = 3 : 7$ .
1 mark
Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :
$\frac{S _{5}}{S _{21}} = \frac{\frac{5}{2} ( 2 a + 4 d )}{\frac{21}{2} ( 2 a + 20 d )} = \frac{5 ( 16 d + 4 d )}{21 ( 16 d + 20 d )} = \frac{5 \times 20 d}{21 \times 36 d}$
1 mark
$= \frac{100}{756} = \frac{25}{189}$
So $S_{5} : S_{21} = 25 : 189$ .
1 mark

Answer: $a_{5} : a_{21} = 3 : 7$ ; $S_{5} : S_{21} = 25 : 189$

Where the marks are

Writes $\frac{a + 10 d}{a + 16 d} = \frac{3}{4}$ (1 mark)
Obtains $a = 8 d$ (1 mark)
Finds $a_{5} : a_{21} = 3 : 7$ (1 mark)
Correct sum-formula expressions for $S_{5}$ and $S_{21}$ with $a = 8 d$ substituted (1 mark)
Simplifies to $S_{5} : S_{21} = 25 : 189$ (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 paper2023 board paper, set 30/1/1 (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 2023: question 35(a)

Hints

Step-by-step solution

Revise the chapter

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