Find the sum of the first 15 terms of the A.P.: 1/15, 1/12, 1/10, …

MathsCBSE Class 10 · Basic · Board exam 2023, set 430/5/1 · 2 marks · Very short answer

CBSE Class 10 Mathematics Basic 2023: question 25

Set in the CBSE Class 10 board exam of 2023, paper 430/5/1, section B, question 25, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q25 · 2 marks · Very short answerArithmetic Progressions

Find the sum of the first 15 terms of the A.P.: $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \dots$

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/5/1.

Hints

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

Hint 1

Find the common difference by subtracting the first term from the second (use LCM 60).

Hint 2

Use $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ with n = 15.

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.

$a = \frac{1}{15}$ , $d = \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}$
½ mark
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow S_{15} = \frac{15}{2} [\frac{2}{15} + 14 \times \frac{1}{60}]$
1 mark
$= \frac{15}{2} [\frac{8}{60} + \frac{14}{60}] = \frac{15}{2} \times \frac{22}{60} = \frac{11}{4}$
½ mark

Answer: $S_{15} = \frac{11}{4}$

Where the marks are

Finds $d = \frac{1}{60}$ (½ mark)
Correct substitution in $S_{n}$ formula (1 mark)
$S_{15} = \frac{11}{4}$ (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)

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 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2023: question 25

Hints

Step-by-step solution

Revise the chapter

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