Chapter 5:Arithmetic Progressions

Class 10 Case based problem of Chapter 5 A.P. 1

Class 10 Case based problem of Chapter 5 A.P. 2

Case Based :4

Sumit’s father had to undergo knee surgery as he was suffering from lot of pain and unable to walk properly. The physiotheropist asked him to return to his jogging program slowly. He suggested jogging for 10 minutes daily for the first week and thereafter increase the jogging time by 6 minutes.

(A) What will be his jogging time after 8 weeks ?

(a) 46 minutes (b) 52 minutes

(B) In which week will his jogging time be 70 minutes per day ?

(a) $8^{th}$ week (b) $10^{th}$ week

(C) Which of the following cannot be the $n^{th}$ term of an A.P. ?

(a) 5 + n (b) 2n + 3

(D) The value of k so that $k^2 + 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4$ are the three consecutive terms of an A.P. is:

(a) 0 (b) -1

(E) Which term of the A.P. : 53, 48, 43, . . . is the first negative term ?

(a) $10^{th}$ (b) $12^{th}$

Solution:

(A) Answer (b) 52 minutes

Explanation: A.P. 10, 16, 22, . . .

a = 1, d = 6

$a_8 = a + 7d$

$= 10 + 7\times 6$

$= 10 + 42$

$= 52$ minutes

(B) Answer (c) $11^{th}$ week

Explanation: $a_n = 70$ minutes

$a_n = a + (n-1)d$

$\Rightarrow 70 = 10 + (n-1)\times 6$

$\Rightarrow 60 = (n-1)\times 6$

$\Rightarrow 10 = n-1$

$\Rightarrow n = 11$

Jogging time will be 70 minutes = $11^{th}$ week

(C) Answer (d) $2n^2+7$ .

Explanation: $n^{th}$ term of any A.P. can not be in quadratic form

Hence, $2n^2+7$ cannot be the $n^{th}$ term of A.P.

(D) Answer (a) 0

Explanation: Since, $k^2 + 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4$ are in A.P.

So, $(2 k^2 + 3k + 6) - (k^2 + 4k + 8) = (3k^2 + 4k + 4) - (2k^2 + 3k + 6)$

$\Rightarrow 2k^2 + 3k + 6- k^2 - 4k -8 = 3k^2 + 4k +4 -2k^2 - 3k - 6$

$\Rightarrow k^2 - k -2 = k^2 + k - 2$

$\Rightarrow -2 k = 0$

$\Rightarrow k = 0$

(E) Answer (b) $12^{th}$

Explanation: A.P. : 53, 48, 43, . . .

a = 53, d = 48 – 53 = -5

Then, $a_n < 0$

$a + (n-1)d < 0$

$\Rightarrow 53 + (n-1)\times -5 < 0$

$\Rightarrow (n-1)\times (-5) < -53$

$\Rightarrow n-1 > \frac{53}{5}$

$\Rightarrow n > 10.6 + 1$

$\Rightarrow n > 11.6$

Hence, n = 12

Therefore, $12^{th}$ term of given A.P. is the first negative term.

Case Based : 5

Kaashvi has a piggy bank made of clay when she has been saving part of her pocket money . She saves ₹ 12 in the first month, ₹ 15 in the second month, ₹ 18 in the third month and continues to save in this manner.

(A) Which of the following is true about the monthly saving of Kaashvi ?

(a) It does not form an A.P.

(b) It forms an A.P. with a = 12, d = 3

(d) It forms an A.P. with a = 12, d = -3

(B) In which month will Kaashvi be able to save ₹ 72 ?

(a) $18^{th}$ (b) $20^{th}$

(C) How much amount did Kaashvi save in the $13^{th}$ month ?

(a) ₹ 57 (b) ₹ 54

(D) The third term of an A.P. is 1 and the sixth term is -11. The $15^{th}$ term of the A.P. is :

(a) -47 (b) -51

(E) The ratio of the $7^{th}$ to the $3^{rd}$ terms of an A.P. is 12 : 5. Then the ratio of its $13^{th}$ to $4^{th}$ terms is :

(a) 3 : 10 (b) 10 : 3

Solution:

(A) Answer (b) It forms an A.P. with a = 12, d = 3

Explanation: As the monthly savings of Kaashvi are ₹ 12, 15, 18 . . .

They form an A.P. with a = 12 and d = 3.

(B) Answer (c) $21^{st}$

Explanation: $a_n =72, a = 12, d = 3$

$a_n = a + (n-1)d$

$\Rightarrow 72 = 12 + (n-1)\times 3$

$\Rightarrow 60 = (n-1)\times 3$

$\Rightarrow 20 = n - 1$

$\Rightarrow n = 21$

Therefore, n = 21

(C) Answer (d) ₹ 48

Explanation: $n = 13, a = 12, d = 3$

$a_n = a + (n-1)\times d$

$\Rightarrow a_{13} = 12 + (13 - 1)\times 3$

$\Rightarrow a_{13} = 12 + 36$

$= 48$

Kaashvi will save in $13^{th}$ month = ₹ 48

(D) Answer (a) -47

Explanation: Given that, $a_{3} = 1$ and $a_{6} = -11$

$a_{3} = a + 2d = 1$ . . . . (i)

$a_{6} = a + 5d = -11$ . . . . . .(ii)

subtracting (ii) to (i)

$(a + 5d) - (a + 2d) = -11 - 1$

$\Rightarrow a + 5d - a - 2d = -12$

$\Rightarrow 3d = -12$

$\Rightarrow d = -4$

Putting in equation (i)

$a + 2d = 1$

$\Rightarrow a + 2(-4) = 1$

$\Rightarrow a - 8 =1$

$\Rightarrow a = 9$

Therefore, $a_{15} = a + 14 d$

$= 9 + 14(-4)$

$= 9 - 56$

$= -47$

The $15^{th}$ term of A.P. = -47

(E) Answer (b) 10 : 3

Explanation: Given that $a_7:a_3 = 12:5$

$\dfrac{a_7}{a_3} = \dfrac{12}{5}$

$\Rightarrow \dfrac{a + 6d}{a + 2d}=\dfrac{12}{5}$

$\Rightarrow 12a + 24d = 5a + 30d$

$\Rightarrow 7a = 6d$

$\Rightarrow d = \frac{7a}{6}$

Now,

$\dfrac{a_{13}}{a_4} =\dfrac{a + 12d}{a + 3d}$

$\Rightarrow \dfrac{a_{13}}{a_4} = \dfrac{a + 12\times \frac{7a}{6}}{a + 3\times \frac{7a}{6}}$

$\Rightarrow \dfrac{a_{13}}{a_4} = \dfrac{a + 14a}{a + \frac{7a}{2}}$

$\Rightarrow \dfrac{a_{13}}{a_4} = \dfrac{15a}{\frac{9a}{2}}$

$\Rightarrow \dfrac{a_{13}}{a_4} = \dfrac{30a}{9 a}$

$\Rightarrow \dfrac{a_{13}}{a_4} = \dfrac{10}{3}$

Hence, Ratio between $13^{th}$ and $4^{th}$ term = 10 : 3

Case Based : 6

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice. It takes him 2 seconds less. He wants to do in 31 seconds.

(A) Which of the following terms are in A.P. for the given situtation ?

(a) 51, 53, 55 . . . . (b) 51, 49, 47 . . .

(B) What is the minimum number of days ?

(a) 10 (b) 12

(C) Which of the following term is not in the A.P. of the above given situtation.

(a) 41 (b) 30

(D) If $n^{th}$ term of an A.P. is given by $a_n = 2n + 3$ , Then common difference of an A.P. is