CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set 03

Case Study Based Questions

I. Your friend Veer wants to participate in a 200 m race. Presently, he can run 200 m in 51 seconds and during each day practice it takes him 2 seconds less. He wants to do in 31 seconds.

Question. Which of the following terms are in AP for the given situation?
(a) 51, 53, 55, ...
(b) 51, 49, 47, ...
(c) –51, –53, –55, ...
(d) 51, 55, 59, …
Answer : B

Question. What is the minimum number of days he needs to practice till his goal is achieved?
(a) 10
(b) 12
(c) 11
(d) 9
Answer : C

Question. Which of the following term is not in the AP of the above given situation?
(a) 41
(b) 30
(c) 37
(d) 39
Answer : B

Question. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is
(a) 2
(b) 3
(c) 5
(d) 1
Answer : A

Question. The value of x, for which 2x, x + 10, 3x + 2 are three consecutive terms of an AP is
(a) 6
(b) – 6
(c) 18
(d) –18
Answer : A
 

II. India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Question. The production during first year is
(a) 3000 TV sets
(b) 5000 TV sets
(c) 7000 TV sets
(d) 10000 TV sets
Answer : B

Question. The production during 8th year is
(a) 10500
(b) 11900
(c) 12500
(d) 20400
Answer : D

Question. The production during first 3 years is
(a) 12800
(b) 19300
(c) 21600
(d) 25200
Answer : C

Question. In which year, the production is 29,200?
(a) 10th year
(b) 12th year
(c) 15th year
(d) 18th year
Answer : B

Question. The difference of the production during 7th year and 4th year is
(a) 6600
(b) 6800
(c) 5400
(d) 7200
Answer : A

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): Common difference of the AP: –5, –1, 3, 7, ... is 4.
Reason (R): Common difference of the AP : a, a + d, a + 2d, ... is given by d = 2^nd term – 1^st term.
Answer : A

Question. Assertion (A): If n^th term of an AP is 7 – 4n, then its common difference is – 4.
Reason (R): Common difference of an AP is given by d = a_n+1 – a_n.
Answer : A

Question. Assertion (A): Common difference of an AP in which a₂₁ – a₇ = 84 is 14.
Reason (R): n^th term of an AP is given by a_n = a + (n – 1) d.
Answer : D

Very Short Answer Type Questions

Question. If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20^th term.
Answer : 200

Question. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
Answer : 16^th term

Question. Find the sum of the first 15 multiples of 8.
Answer : 960

Question. Determine the AP whose third term is 16 and the 7^th term exceeds the 5^th term by 12.
Answer : 4, 10, 16, 22, . . .

Question. Find the number of terms in the AP : 7, 13, 19, . . . , 205
Answer : 34

Question. The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Answer : –13, –8, –3

Arithmetic Progression

Question. If 5 times the 5^th term of an AP is equal to 10 times the 10th term, show that its 15th term is zero.
Answer : Let 1st term = a and common difference = d.
a₅ = a + 4d, a₁₀ = a + 9d
According to the question, 5 × a₅ = 10 × a₁₀
=> 5(a + 4d) = 10(a + 9d)
=> 5a + 20d = 10a + 90d a = -14d
Now, a15 = a + 14d
=> a₁₅ = -14d + 14d = 0. 

Very Short Answer Type Questions (1 Mark)

Question. The \(n\)th term of an AP is \(7-4n\). Find its common difference.
Answer: -4

Question. Which term of the AP \(21, 18, 15, \ldots\), is zero?
Answer: 8

Question. For what value of \(p\), are \(2p + 1, 13, 5p - 3\) three consecutive terms of an AP?
Answer: 4

Question. If \(a_n = \frac{n(n-3)}{n+4}\), then find \(18^{\text{th}}\) term of this sequence.
Answer: \(\frac{135}{11}\)

Question. If the sum of first \(m\) terms of an AP is \(2m^2 + 3m\), then what is its second term?
Answer: 9

Question. If the sum of first \(p\) terms of an AP is \(ap^2 + bp\), find its common difference.
Answer: \(2a\)

Short Answer Type Questions I (2 Marks)

Question. Is \(-150\) a term of the AP \(17, 12, 7, 2, \ldots\)?
Answer: No, since \(n = \frac{172}{5}\) is not an integer.

Question. In an AP, the sum of first \(n\) terms is \(\frac{5n^2}{2} + \frac{3n}{2}\). Find its \(20^{\text{th}}\) term.
Answer: 99

Question. Find the common difference of an AP whose first term is 4, the last term is 49 and the sum of all its terms is 265.
Answer: 5

Question. The sum of three numbers of an AP is 27 and their product is 405. Find the numbers.
Answer: 3, 9, 15 and 15, 9, 3

Question. Find \(k\), if the given value of \(x\) is the \(k^{\text{th}}\) term of the given AP: \(25, 50, 75, 100, \ldots, x = 1000\)
Answer: 40

Short Answer Type Questions II (3 Marks)

Question. If \(m\) times the \(m^{\text{th}}\) term of an AP is equal to \(n\) times its \(n^{\text{th}}\) term, find the \((m + n)^{\text{th}}\) term of the AP.
Answer: 0

Question. If \(9^{\text{th}}\) term of an AP is zero, prove that its \(29^{\text{th}}\) term is double of its \(19^{\text{th}}\) term.
Answer: Let \(a\) be the first term and \(d\) be the common difference. Given \(a_9 = 0 \implies a + 8d = 0 \implies a = -8d\).
The \(19^{\text{th}}\) term is \(a_{19} = a + 18d = -8d + 18d = 10d\).
The \(29^{\text{th}}\) term is \(a_{29} = a + 28d = -8d + 28d = 20d\).
Clearly, \(a_{29} = 2 \cdot a_{19}\). Hence Proved.

Question. Which term of the progression \(19, 18\frac{1}{5}, 17\frac{2}{5}, \ldots\) is the first negative term?
Answer: \(25^{\text{th}}\) term

Question. If the \(p^{\text{th}}\), \(q^{\text{th}}\), \(r^{\text{th}}\) terms of an AP be \(x, y, z\) respectively, show that \(x(q - r) + y(r - p) + z(p - q) = 0\).
Answer: Let the first term of the AP be \(a\) and the common difference be \(d\). Then,
\(x = a + (p - 1)d\)
\(y = a + (q - 1)d\)
\(z = a + (r - 1)d\)
Substituting these values in the expression:
\(x(q - r) + y(r - p) + z(p - q) = [a + (p - 1)d](q - r) + [a + (q - 1)d](r - p) + [a + (r - 1)d](p - q)\)
\(= a[(q - r) + (r - p) + (p - q)] + d[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]\)
\(= a(0) + d[(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)]\)
\(= 0 + d(0) = 0\). Hence Proved.

Question. How many terms of the AP \(-6, -\frac{11}{2}, -5, \ldots\) are needed to give the sum \(-25\)? Explain the double answer.
Answer: \(n = 5, 20\). The double answer occurs because the sum of the terms from the \(6^{\text{th}}\) term to the \(20^{\text{th}}\) term is zero, as the terms change from negative to positive.

Question. If the \(p^{\text{th}}\) term of an AP is \(\frac{1}{q}\) and the \(q^{\text{th}}\) term is \(\frac{1}{p}\), show that the sum of \(pq\) terms is \(\frac{1}{2}(pq + 1)\).
Answer: Let \(a\) be the first term and \(d\) be the common difference.
\(a_p = a + (p - 1)d = \frac{1}{q}\) — (1)
\(a_q = a + (q - 1)d = \frac{1}{p}\) — (2)
Subtracting (2) from (1):
\((p - q)d = \frac{1}{q} - \frac{1}{p} = \frac{p - q}{pq} \implies d = \frac{1}{pq}\).
Substituting \(d = \frac{1}{pq}\) into (1):
\(a + \frac{p - 1}{pq} = \frac{1}{q} \implies a = \frac{1}{pq}\).
Now, the sum of first \(pq\) terms is:
\(S_{pq} = \frac{pq}{2}\left[2a + (pq - 1)d\right] = \frac{pq}{2}\left[\frac{2}{pq} + \frac{pq - 1}{pq}\right] = \frac{1}{2}(pq + 1)\). Hence Proved.

Long Answer Type Questions (4 Marks)

Question. If the sum of first 4 terms of an AP is 40 and that of first 14 terms is 280, find the sum of its first \(n\) terms.
Answer: \(n(n + 6)\)

Question. The first and the last terms of an AP are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum?
Answer: \(n = 39\), Sum = \(6981\)

Question. How many multiples of 4 lie between 10 and 250? Also find their sum.
Answer: 60 multiples, Sum = \(7800\)

Question. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Answer: 2

Question. Find the sum of the integers between 100 and 200 that are
i) divisible by 9
ii) not divisible by 9
Answer: i) 1683, ii) 13167

Question. The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2m. The flags are stored at the position of the middle most flag. Ruchi was given responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing the job and returning back to collect her books. What is the maximum distance she travelled carrying a flag?
Answer: Distance covered in completing the job = 728m, Maximum distance travelled carrying a flag = 26m

Question. If \(S_1, S_2, S_3\) are the sum of \(n\) terms of three APs, the first term of each being unity and the respective common difference being 1, 2, 3; prove that \(S_1 + S_3 = 2S_2\).
Answer: Let the sums of \(n\) terms of the three APs be \(S_1\), \(S_2\), and \(S_3\).
With first term \(a = 1\) for all three APs, and common differences \(d_1 = 1\), \(d_2 = 2\), \(d_3 = 3\):
\(S_1 = \frac{n}{2}\left[2(1) + (n - 1)(1)\right] = \frac{n(n + 1)}{2}\)
\(S_2 = \frac{n}{2}\left[2(1) + (n - 1)(2)\right] = \frac{n}{2}[2n] = n^2\)
\(S_3 = \frac{n}{2}\left[2(1) + (n - 1)(3)\right] = \frac{n(3n - 1)}{2}\)
Now, adding \(S_1\) and \(S_3\):
\(S_1 + S_3 = \frac{n(n + 1)}{2} + \frac{n(3n - 1)}{2} = \frac{n(n + 1 + 3n - 1)}{2} = \frac{n(4n)}{2} = 2n^2 = 2S_2\). Hence Proved.