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This Class 10 Mathematics topic applies Arithmetic Progressions (AP) to real-life word problems. Students learn to identify the first term, common difference, number of terms, and required sum from situations involving regular increases or decreases, such as savings, seating arrangements, wages, distances, and patterns. They practise translating statements into AP terms, selecting suitable formulas, solving step by step, and checking whether the answer fits the original context.
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Hard · Level 71 · arithmetic progression,ap word problems,nth term,linear sequence,class 10 mathematicsView options
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Hard · Level 71 · ap,word-problem,work,bricksView options
(2240)
(2280)
(2320)
(2360)
Hard · Level 71 · ap,word-problem,donation,targetView options
(8)
(9)
(10)
(11)
Hard · Level 71 · arithmetic progression, ap word problems, staircase tiles, decreasing sequence, nth termView options
10th row
11th row
12th row
13th row
Hard · Level 71 · ap,word-problem,taxi,fareView options
(1848)
(1908)
(1968)
(2028)
Hard · Level 71 · ap,word-problem,matchstick,patternView options
(122)
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(130)
(134)
Hard · Level 71 · ap,word-problem,fund,totalView options
(6940)
(7040)
(7140)
(7240)
Hard · Level 71 · ap,word-problem,reservoir,totalView options
(12250)
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(12750)
(13000)
Hard · Level 71 · ap,word-problem,salary,nth-termView options
(42000)
(43500)
(45000)
(46500)
Hard · Level 71 · ap,word-problem,parking,capacityView options
(1683)
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(1719)
(1737)
Hard · Level 71 · ap,word-problem,pottery,totalView options
(921)
(951)
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(1011)
Hard · Level 71 · ap,word-problem,admissions,targetView options
(18)
(19)
(20)
(21)
Hard · Level 71 · ap,word-problem,machine,nth-termView options
(224)
(236)
(248)
(260)
Hard · Level 71 · ap,word-problem,construction,hoursView options
(500)
(510)
(520)
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Hard · Level 71 · ap,word-problem,installment,totalView options
(50000)
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Medium · Level 71 · arithmetic progression,AP word problem,sum of terms,Word problems based on APs,Arithmetic Progressions (AP),arithmetic progressions ap,Mathematics,Class 10 MCQView options
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Hard · Level 71 · ap,word-problem,sales,decreaseView options
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Hard · Level 71 · arithmetic progression, ap word problems, nth term, class 10 mathematics, linear equationsView options
13
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Hard · Level 71 · ap,word-problem,pole-distance,totalView options
(930)
(948)
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(984)
Hard · Level 71 · ap,word-problem,gym,membersView options
(690)
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(726)
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Question 1HardLevel 71
In a garden the first row has (16) plants and each next row has (6) more plants. Which row will have (112) plants?
Correct answer: C
The numbers of plants form an arithmetic progression with first term 16 and common difference 6. Thus, the number of plants in the nth row is \(a_n=16+(n-1)\times6\). On putting \(16+(n-1)\times6=112\), we get \(n-1=16\), so \(n=17\). Therefore, the 17th row has 112 plants. The 16th row would have only 106 plants. Exam tip: in AP word problems, identify the first term and common difference, then equate the target value to \(a_n\).
In a donation campaign (150) rupees are received on the first day and (75) rupees more each next day. How many days will it take to collect (4875) rupees?
Correct answer: C
The donation amounts are (150,225,300,\ldots) and (S_n=4875) gives (n=10). Exam tip: form an (S_n) equation for the target total.
A staircase has (110) tiles in the bottom row and (5) fewer tiles in each upper row. Which row will have (55) tiles?
Correct answer: C
The numbers of tiles, counted from the bottom, form a decreasing AP with first term 110 and common difference -5. Thus, the number of tiles in the nth row is \(110+(n-1)(-5)\). Setting this equal to 55 gives \(110-5(n-1)=55\), so \(n-1=11\) and \(n=12\). Therefore, the 12th row has 55 tiles. The 11th row has 60 tiles, so it is a close but incorrect option. Exam tip: use a negative common difference when the terms decrease upwards.
In a matchstick pattern the first figure has (10) matchsticks and each next figure has (8) more matchsticks. How many matchsticks are in the (16)th figure?
Correct answer: C
The matchstick numbers are (10,18,26,\ldots) and (a_{16}=130). Exam tip: take the figure number as the term number.
A reservoir has (1200) litres of water on the first day and (50) litres less each next day. What is the total observed water amount over the first (15) days?
Correct answer: C
The amounts form the decreasing AP (1200,1150,1100,\ldots) and (S_{15}=12750). Exam tip: keep (d) negative for decreasing quantities.
A machine produces (80) units in the first hour and (12) more units each next hour. What is the production in the (14)th hour?
Correct answer: B
The hourly production forms an arithmetic progression because the machine adds the same number of units, 12, every hour. Here the first term is 80, the common difference is 12, and the 14th hour means the 14th term. The term formula is \\(a_n=a+(n-1)d\\), so \\(a_{14}=80+(14-1)12=80+156=236\\).
Therefore, option B, 236, is correct. It is important not to multiply 14 directly by 12, because the first hour already starts at 80 and only the next 13 hours add an increase. The sequence begins 80, 92, 104 and continues with a constant difference of 12. Treating the hour number as the term number gives the required production.
A theatre has 928 seats in 16 rows. If each next row has 4 more seats, how many seats were in the first row?
Correct answer: C
The governing concept is the sum of the first n terms of an arithmetic progression. Here the number of seats forms an AP with n = 16, common difference d = 4, first term a, and total S₁₆ = 928. Use Sₙ = n/2[2a + (n−1)d]. Thus 928 = 16/2[2a + 15(4)] = 8(2a + 60). Dividing by 8 gives 116 = 2a + 60, so 2a = 56 and a = 28. Therefore option C is correct. Options A, B and D do not satisfy the given total when the seats increase by four in each successive row.
A student reads (25) pages in the first month and (15) pages more each next month. In which month will the student read (235) pages?
Correct answer: C
The numbers of pages form an arithmetic progression with first term \(a=25\) and common difference \(d=15\). The number of pages read in the \(n\)th month is \(a_n=a+(n-1)d\). So, \(25+(n-1)\times15=235\). This gives \((n-1)\times15=210\), hence \(n-1=14\) and \(n=15\). Therefore, the student reads 235 pages in the 15th month. In the 14th month, the student would read only 220 pages. Exam tip: for a specified term of an AP, use \(a_n=a+(n-1)d\).
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