How many terms of the arithmetic progression 45, 39, 33, … must be taken so that their sum is 180 ? Explain the double answer.

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 32

Set in the CBSE Class 10 board exam of 2023, paper 30/2/1, section D, question 32, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 5 marks · Long answerArithmetic Progressions

How many terms of the arithmetic progression $45, 39, 33, \dots$ must be taken so that their sum is $180$ ? Explain the double answer.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/2/1.

Hints

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

Hint 1

Write down $a$ and $d$ , then use the sum formula $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

Hint 2

You will get a quadratic in $n$ with two positive roots. Check what the terms between them add up to.

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.

Here $a = 45$ , $d = 39 - 45 = - 6$ and $S_{n} = 180$ .
½ mark
Using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ :
$180 = \frac{n}{2} [90 + (n - 1) (- 6)] = \frac{n}{2} [96 - 6 n]$
1 mark
$360 = 96 n - 6 n^{2} \Rightarrow 6 n^{2} - 96 n + 360 = 0$
1 mark
Dividing by 6: $n^{2} - 16 n + 60 = 0 \Rightarrow (n - 6) (n - 10) = 0$
1 mark
$n = 6$ or $n = 10$
1 mark
Both answers are valid. Explanation: $a_{7} = 45 + 6 (- 6) = 9$ , $a_{8} = 3$ , $a_{9} = - 3$ , $a_{10} = - 9$ . The positive and negative terms cancel, so $a_{7} + a_{8} + a_{9} + a_{10} = 0$ . Hence the sum of the first 10 terms equals the sum of the first 6 terms, i.e. 180.
½ mark

Answer: 6 terms or 10 terms. The 7th to 10th terms add up to 0, so both sums are 180.

Where the marks are

Identifies $a = 45$ , $d = - 6$ (½ mark)
Substitutes into the $S_{n}$ formula (1 mark)
Forms the quadratic $6 n^{2} - 96 n + 360 = 0$ (1 mark)
Factorises $n^{2} - 16 n + 60 = 0$ (1 mark)
Gets $n = 6, 10$ (1 mark)
Explains that the 7th to 10th terms sum to zero (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)
Topic: Sum of First n Terms of an AP

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 30/2/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 32

Hints

Step-by-step solution

Revise the chapter

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