CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set 16

Question. In an A.P., \( S_1 = 6, S_7 = 105 \), then find \( S_n : S_{n-3} \).
Answer: Given \( S_1 = 6 \), which means the first term \( a = 6 \).
Also, \( S_7 = 105 \)
\( \implies \frac{7}{2}[2a + (7 - 1)d] = 105 \)
\( \implies \frac{7}{2}[2(6) + 6d] = 105 \)
\( \implies 7(6 + 3d) = 105 \)
\( \implies 6 + 3d = 15 \)
\( \implies 3d = 9 \)
\( \implies d = 3 \)
Now, \( S_n : S_{n-3} = \frac{\frac{n}{2}[2(6) + (n-1)3]}{\frac{n-3}{2}[2(6) + (n-3-1)3]} \)
\( \implies S_n : S_{n-3} = \frac{n[12 + 3n - 3]}{(n-3)[12 + 3n - 12]} \)
\( \implies S_n : S_{n-3} = \frac{n(3n + 9)}{(n-3)(3n)} \)
\( \implies S_n : S_{n-3} = \frac{3n(n + 3)}{3n(n - 3)} = \frac{n + 3}{n - 3} \).
Thus, the ratio is \( (n + 3) : (n - 3) \).

Question. Let \( S_1, S_2, S_3 \) be the sums of \( n \) terms of three series in A.P., the first term of each being 1 and the common differences 1, 2, 3 respectively. If \( S_1 + S_3 = \lambda S_2 \), then find the value of \( \lambda \)?
Answer: \( S_1 = \frac{n}{2}[2(1) + (n-1)1] = \frac{n(n+1)}{2} \)
\( S_2 = \frac{n}{2}[2(1) + (n-1)2] = \frac{n(2n)}{2} = n^2 \)
\( S_3 = \frac{n}{2}[2(1) + (n-1)3] = \frac{n(3n-1)}{2} \)
Given \( S_1 + S_3 = \lambda S_2 \)
\( \implies \frac{n(n+1)}{2} + \frac{n(3n-1)}{2} = \lambda n^2 \)
\( \implies \frac{n(n + 1 + 3n - 1)}{2} = \lambda n^2 \)
\( \implies \frac{n(4n)}{2} = \lambda n^2 \)
\( \implies 2n^2 = \lambda n^2 \)
\( \implies \lambda = 2 \).

Question. Find the 12th term from the end of the arithmetic progression: \( 3, 5, 7, 9, \dots, 201 \)?
Answer: Here, last term \( L = 201 \) and common difference \( d = 5 - 3 = 2 \).
The \( n \)th term from the end is given by \( L - (n-1)d \).
\( \text{12th term from the end} = 201 - (12 - 1) \times 2 \)
\( \implies 201 - 11 \times 2 \)
\( \implies 201 - 22 = 179 \).

Question. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second terms by 6, find three terms.
Answer: Let the three terms be \( a - d, a, a + d \).
Sum \( = (a - d) + a + (a + d) = 21 \)
\( \implies 3a = 21 \implies a = 7 \).
Product of 1st and 3rd exceeds 2nd by 6:
\( (a - d)(a + d) = a + 6 \)
\( \implies a^2 - d^2 = a + 6 \)
\( \implies 7^2 - d^2 = 7 + 6 \)
\( \implies 49 - d^2 = 13 \)
\( \implies d^2 = 36 \implies d = \pm 6 \).
Taking \( d = 6 \), the terms are \( 7 - 6, 7, 7 + 6 \), i.e., \( 1, 7, 13 \).
Taking \( d = -6 \), the terms are \( 13, 7, 1 \).

Question. The sum of three numbers in an A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Answer: Let the numbers be \( a - d, a, a + d \).
Sum \( = 3a = 12 \implies a = 4 \).
Sum of cubes \( = (a - d)^3 + a^3 + (a + d)^3 = 288 \)
\( \implies (a^3 - 3a^2d + 3ad^2 - d^3) + a^3 + (a^3 + 3a^2d + 3ad^2 + d^3) = 288 \)
\( \implies 3a^3 + 6ad^2 = 288 \)
\( \implies 3(4)^3 + 6(4)d^2 = 288 \)
\( \implies 192 + 24d^2 = 288 \)
\( \implies 24d^2 = 96 \)
\( \implies d^2 = 4 \implies d = \pm 2 \).
The numbers are \( 2, 4, 6 \).

Question. Two A.P.’s have the same common difference. If the first terms of the two A.P’s are 3 and 8 respectively, find the difference between their sum of first 30 terms.
Answer: Let the common difference be \( d \).
For the first AP: \( a = 3 \).
\( S_{30} = \frac{30}{2}[2(3) + 29d] = 15(6 + 29d) = 90 + 435d \).
For the second AP: \( A = 8 \).
\( S'_{30} = \frac{30}{2}[2(8) + 29d] = 15(16 + 29d) = 240 + 435d \).
Difference \( = S'_{30} - S_{30} = (240 + 435d) - (90 + 435d) = 150 \).

Question. If \( x, y, z \) are in A.P., show that \( (x + 2y - z) (2y + z - x) (z + x - y) = 4xyz \).
Answer: Since \( x, y, z \) are in AP, \( 2y = x + z \).
LHS \( = (x + 2y - z) (2y + z - x) (z + x - y) \)
Substituting \( 2y = x + z \) in the first two brackets and \( x + z = 2y \) in the third:
\( \implies (x + x + z - z) (x + z + z - x) (2y - y) \)
\( \implies (2x) (2z) (y) = 4xyz = \) RHS. Hence Proved.

Question. A ball rolling on an inclined plane covers 30m during the 1st, 27 m during the 2nd and 24 during the 3rd second and so on. How much distance does it travel during the 10th second? Also find the total distance covered in 10 seconds.
Answer: Distances form an AP: \( 30, 27, 24, \dots \)
Here \( a = 30 \) and \( d = 27 - 30 = -3 \).
Distance in 10th second \( = a_{10} = a + 9d = 30 + 9(-3) = 30 - 27 = 3 \) m.
Total distance in 10 seconds \( = S_{10} = \frac{10}{2}[2(30) + 9(-3)] \)
\( \implies 5[60 - 27] = 5 \times 33 = 165 \) m.

Question. Which term of the A.P., \( -9, -8.25, -7.5, \dots \) is its first positive term?
Answer: Here \( a = -9 \) and \( d = -8.25 - (-9) = 0.75 \).
Let the \( n \)th term be the first positive term, so \( a_n > 0 \).
\( \implies a + (n-1)d > 0 \)
\( \implies -9 + (n-1)(0.75) > 0 \)
\( \implies 0.75(n-1) > 9 \)
\( \implies n - 1 > \frac{9}{0.75} = 12 \)
\( \implies n > 13 \).
Thus, the 14th term is the first positive term.

Question. An A.P. consists of 37 terms. The sum of three middle most terms is 225 and the sum of the last three terms is 429. Find the A.P.
Answer: With 37 terms, the middle term is the \( \frac{37+1}{2} = 19 \)th term.
The three middle-most terms are \( a_{18}, a_{19}, a_{20} \).
Sum \( = a_{18} + a_{19} + a_{20} = 225 \implies 3a_{19} = 225 \implies a + 18d = 75 \) --- (i)
Last three terms are \( a_{35}, a_{36}, a_{37} \).
Sum \( = (a + 34d) + (a + 35d) + (a + 36d) = 429 \implies 3a + 105d = 429 \implies a + 35d = 143 \) --- (ii)
Subtracting (i) from (ii): \( 17d = 68 \implies d = 4 \).
From (i), \( a + 18(4) = 75 \implies a + 72 = 75 \implies a = 3 \).
The AP is \( 3, 7, 11, \dots \)

Question. If nth terms of the sequences \( 3, 10, 17, \dots \) and \( 63, 65, 67, \dots \) are equal, then find the value of \( n \)?
Answer: For the first sequence: \( a_n = 3 + (n-1)7 = 7n - 4 \).
For the second sequence: \( A_n = 63 + (n-1)2 = 2n + 61 \).
Given \( a_n = A_n \)
\( \implies 7n - 4 = 2n + 61 \)
\( \implies 5n = 65 \)
\( \implies n = 13 \).

Question. A body falls 16 metres in the first second of its motion, 48 metres in the second, 80 metres in the third and so on. How long will it take to fall 4096 metres?
Answer: Distances form an AP: \( 16, 48, 80, \dots \) where \( a = 16, d = 32 \).
Let it take \( n \) seconds.
\( S_n = \frac{n}{2}[2a + (n-1)d] = 4096 \)
\( \implies \frac{n}{2}[2(16) + (n-1)32] = 4096 \)
\( \implies n[16 + 16n - 16] = 4096 \)
\( \implies 16n^2 = 4096 \)
\( \implies n^2 = 256 \implies n = 16 \).
It will take 16 seconds.

Question. If the sum of the series \( (x + 1) + (x + 4) + (x + 7) + \dots + (x + 28) = 155 \), then find the value of \( x \)?
Answer: The constant terms are \( 1, 4, 7, \dots, 28 \).
\( a = 1, d = 3, a_n = 28 \)
\( \implies 28 = 1 + (n-1)3 \implies n - 1 = 9 \implies n = 10 \).
The series has 10 terms.
\( \implies 10x + (1 + 4 + 7 + \dots + 28) = 155 \)
\( \implies 10x + \frac{10}{2}(1 + 28) = 155 \)
\( \implies 10x + 5(29) = 155 \)
\( \implies 10x + 145 = 155 \)
\( \implies 10x = 10 \implies x = 1 \).

Question. Find the value of \( [(1 - \frac{1}{n+1}) + (1 - \frac{2}{n+1}) + (1 - \frac{3}{n+1}) + \dots + (1 - \frac{n}{n+1})] \).
Answer: The expression consists of \( n \) terms.
\( = (1 + 1 + \dots \text{ to } n \text{ terms}) - \frac{1}{n+1}(1 + 2 + 3 + \dots + n) \)
\( \implies n - \frac{1}{n+1} \times \frac{n(n+1)}{2} \)
\( \implies n - \frac{n}{2} = \frac{n}{2} \).

Question. Assertion (A): The 6th term the end of the AP \( 5, 2, -1, -4, \dots, -31 \) is \( -16 \).
Reason (R): General term (\( n \)th term) from the beginning is given by \( a_n = a + (n - 1)d \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true but R is not the correct explanation of A.

Question. Assertion (A): 12th term of the AP \( 30, 27, 24 \dots \) is the first negative term of the AP.
Reason (R): Sum of first \( n \) terms of an AP \( a, a + d, a + 2d, \dots \) is given by \( S_n = \frac{n}{2}\{2a + (n - 1)d\} \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true but R is not the correct explanation of A.

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs. 1,18,000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, answer the following:

Question. Find the amount paid by him in 30th instalment.
Answer: Here \( a = 1000, d = 100 \).
Amount in 30th instalment \( = a_{30} = a + 29d \)
\( \implies 1000 + 29(100) = 1000 + 2900 = \text{Rs. } 3900 \).

Question. If total installments are 40, then find the amount paid in the last instalment.
Answer: Last instalment is the 40th instalment.
\( a_{40} = a + 39d \)
\( \implies 1000 + 39(100) = 1000 + 3900 = \text{Rs. } 4900 \).

Question. Find the amount paid by him in the 30 instalments.
OR
Find the ratio of the 1st installment to the last instalment.
Answer: Total paid in 30 instalments \( = S_{30} = \frac{30}{2}[2a + 29d] \)
\( \implies 15[2(1000) + 29(100)] \)
\( \implies 15[2000 + 2900] = 15 \times 4900 = \text{Rs. } 73,500 \).
OR
Assuming 40 instalments as the last, ratio \( = a_1 : a_{40} \)
\( \implies 1000 : 4900 = 10 : 49 \).

Question. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \( \frac{1}{4} \) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
(iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
Answer: (i) Given, \( a_1 = \text{Rs. } 15 \), \( a_2 = \text{Rs. } 15 + \text{Rs. } 8 = \text{Rs. } 23 \), \( a_3 = \text{Rs. } 23 + \text{Rs. } 8 = \text{Rs. } 31 \)
\( \therefore \) List of fares is Rs. 15, Rs. 23, Rs. 31 .....
Now, \( a_2 - a_1 = \text{Rs. } 23 - \text{Rs. } 15 = \text{Rs. } 8 \)
\( a_3 - a_2 = \text{Rs. } 31 - \text{Rs. } 23 = \text{Rs. } 8 \)
Here, \( a_2 - a_1 = a_3 - a_2 \)
Thus, the list forms an AP.
(ii) Let \( a_1 = x \); \( a_2 = x - \frac{1}{4}x = \frac{3}{4}x \);
\( a_3 = \frac{3}{4}x - \frac{1}{4}(\frac{3}{4}x) \)
\( = \frac{3}{4}x - \frac{3}{16}x = \frac{9}{16}x \)
The list of numbers is \( x, \frac{3}{4}x, \frac{9}{16}x \dots \)
\( a_2 - a_1 = \frac{3}{4}x - x = -\frac{1}{4}x \);
\( a_3 - a_2 = \frac{9}{16}x - \frac{3}{4}x = -\frac{3}{16}x \)
Here, \( a_2 - a_1 \neq a_3 - a_2 \)
\( \therefore \) It is not an AP.
(iii) Given, \( a_1 = \text{Rs. } 150 \), \( a_2 = \text{Rs. } 200 \), \( a_3 = \text{Rs. } 250 \)
\( \therefore a_2 - a_1 = \text{Rs. } 200 - \text{Rs. } 150 = \text{Rs. } 50 \)
and \( a_3 - a_2 = \text{Rs. } 250 - \text{Rs. } 200 = \text{Rs. } 50 \)
Here, \( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The list forms an AP.
(iv) \( a_1 = \text{Rs. } 10000 \)
\( a_2 = \text{Rs. } 10000 + \text{Rs. } 10000 \times \frac{8}{100} \)
\( = \text{Rs. } 10000 + \text{Rs. } 800 = \text{Rs. } 10800 \)
\( a_3 = \text{Rs. } 10800 + \text{Rs. } 10800 \times \frac{8}{100} \)
\( = \text{Rs. } 10800 + \text{Rs. } 864 \)
\( = \text{Rs. } 11664 \)
Here, \( a_2 - a_1 = \text{Rs. } 10800 - \text{Rs. } 10000 \)
\( = \text{Rs. } 800 \)
and \( a_3 - a_2 = \text{Rs. } 11664 - \text{Rs. } 10800 = \text{Rs. } 864 \)
\( \because a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) It is not an AP.

Question. Write first four terms of the AP, when the first term \( a \) and the common difference \( d \) are given as follows:
(i) \( a = 10, d = 10 \)
(ii) \( a = -2, d = 0 \)
(iii) \( a = 4, d = -3 \)
(iv) \( a = -1, d = \frac{1}{2} \)
(v) \( a = -1.25, d = -0.25 \)
Answer: (i) Given, \( a = 10, d = 10 \)
\( a_1 = 10, a_2 = 10 + 10 = 20 \)
\( a_3 = 20 + 10 = 30 \);
\( a_4 = 30 + 10 = 40 \)
Thus, the first four terms of the AP are 10, 20, 30 and 40 respectively.
(ii) Given, \( a = -2, d = 0 \)
The first four terms of the AP are \( -2, -2, -2 \) and \( -2 \).
(iii) \( a_1 = 4, d = -3 \)
and \( a_2 = 4 + d = 4 - 3 = 1 \)
\( a_3 = 1 + d = 1 - 3 = -2 \)
and \( a_4 = -2 + d = -2 - 3 = -5 \)
\( \therefore \) The first four terms are 4, 1, \( -2 \) and \( -5 \).
(iv) \( a_1 = -1, d = \frac{1}{2} \)
and \( a_2 = -1 + \frac{1}{2} = -\frac{1}{2} \)
\( a_3 = -\frac{1}{2} + d = -\frac{1}{2} + \frac{1}{2} = 0 \)
and \( a_4 = 0 + d = 0 + \frac{1}{2} = \frac{1}{2} \)
\( \therefore \) The first four terms of the AP are \( -1, -\frac{1}{2}, 0, \frac{1}{2} \)
(v) \( a_1 = -1.25, d = -0.25 \)
and \( a_2 = -1.25 + d = -1.25 - 0.25 = -1.50 \)
\( a_3 = -1.50 + d = -1.50 - 0.25 = -1.75 \)
and \( a_4 = -1.75 + d = -1.75 - 0.25 = -2 \)
\( \therefore \) The first four terms of the AP are \( -1.25, -1.50, -1.75, -2 \).

Question. For the following APs, write the first term and the common difference:
(i) \( 3, 1, -1, -3, \dots \)
(ii) \( -5, -1, 3, 7, \dots \)
(iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3} \dots \)
(iv) \( 0.6, 1.7, 2.8, 3.9, \dots \)
Answer: (i) \( a_1 = 3, a_2 = 1 \)
\( d = a_2 - a_1 = 1 - 3 = -2 \)
where \( a_1 = \text{first term} \)
and \( d = \text{common difference} \)
\( \therefore a_1 = 3, d = -2 \)
(ii) \( a_1 = -5, a_2 = -1 \)
\( \therefore d = a_2 - a_1 = -1 - (-5) = -1 + 5 = 4 \)
So, first term \( a_1 = -5 \), common difference, \( d = 4 \)
(iii) \( a_1 = \frac{1}{3}, a_2 = \frac{5}{3} \)
\( \therefore d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \)
So, first term \( a_1 = \frac{1}{3} \) and common difference \( d = \frac{4}{3} \)
(iv) \( 0.6, 1.7, 2.8, 3.9 \)
Here, \( a_1 = 0.6, a_2 = 1.7 \)
\( \therefore d = a_2 - a_1 = 1.7 - 0.6 = 1.1 \)
So, first term \( a_1 = 0.6 \) and common difference \( d = 1.1 \)

Question. Which of the following are APs? If they form an AP, find the common difference \( d \) and write three more terms.
(i) \( 2, 4, 8, 16, \dots \)
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
(iii) \( -1.2, -3.2, -5.2, -7.2, \dots \)
(iv) \( -10, -6, -2, 2, \dots \)
(v) \( 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots \)
(vi) \( 0.2, 0.22, 0.222, 0.2222, \dots \)
(vii) \( 0, -4, -8, -12, \dots \)
(viii) \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots \)
(ix) \( 1, 3, 9, 27, \dots \)
(x) \( a, 2a, 3a, 4a, \dots \)
(xi) \( a, a^2, a^3, a^4, \dots \)
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \)
(xiv) \( 1^2, 3^2, 5^2, 7^2, \dots \)
(xv) \( 1^2, 5^2, 7^2, 73, \dots \)
Answer: (i) \( 2, 4, 8, 16, \dots \)
Here, \( a_2 - a_1 = 4 - 2 = 2 \), \( a_3 - a_2 = 8 - 4 = 4 \)
\( \because a_2 - a_1 \neq a_3 - a_2 \)
\( \therefore \) It is not an AP.
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)
Here, \( a_2 - a_1 = \frac{5}{2} - 2 = \frac{1}{2} \)
and \( a_3 - a_2 = 3 - \frac{5}{2} = \frac{1}{2} \)
\( \because a_2 - a_1 = a_3 - a_2 \)
The given sequence is an AP.
Here \( a_1 = a = 2, d = \frac{1}{2} \)
\( \therefore a_5 = \frac{7}{2} + \frac{1}{2} = 4 \);
\( a_6 = 4 + \frac{1}{2} = \frac{9}{2} \); \( a_7 = \frac{9}{2} + \frac{1}{2} = 5 \)
(iii) \( -1.2, -3.2, -5.2, -7.2, \dots \)
Here, \( a_2 - a_1 = -3.2 - (-1.2) = -3.2 + 1.2 = -2 \)
\( a_3 - a_2 = -5.2 - (-3.2) = -5.2 + 3.2 = -2 \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Here, \( a_1 = a = -1.2, d = -2 \)
\( \therefore a_5 = -7.2 + (-2) = -9.2 \);
\( a_6 = (-9.2) + (-2) = -11.2 \);
\( a_7 = (-11.2) + (-2) = -13.2 \)
(iv) \( -10, -6, -2, 2, \dots \)
\( a_2 - a_1 = -6 - (-10) = 4 \)
and \( a_3 - a_2 = -2 - (-6) = 4 \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Here, \( a_1 = a = -10, d = 4 \)
\( \therefore a_5 = 2 + 4 = 6 \); \( a_6 = 6 + 4 = 10 \);
\( a_7 = 10 + 4 = 14 \)
(v) \( 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \dots \)
Here, \( a_2 - a_1 = 3 + \sqrt{2} - 3 = \sqrt{2} \)
and \( a_3 - a_2 = 3 + 2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2} \)
\( \therefore a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Now, \( a_1 = 3, d = \sqrt{2} \)
\( \therefore a_5 = 3 + 3\sqrt{2} + \sqrt{2} = 3 + 4\sqrt{2} \);
\( a_6 = 3 + 4\sqrt{2} + \sqrt{2} = 3 + 5\sqrt{2} \)
\( a_7 = 3 + 5\sqrt{2} + \sqrt{2} = 3 + 6\sqrt{2} \)
(vi) \( 0.2, 0.22, 0.222, 0.2222, \dots \)
Here, \( a_2 - a_1 = 0.22 - 0.2 = 0.02 \)
and \( a_3 - a_2 = 0.222 - 0.22 = 0.002 \)
\( a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) The given sequence is not an AP.
(vii) \( 0, -4, -8, -12, \dots \)
Here, \( a_2 - a_1 = -4 - 0 = -4 \)
and \( a_3 - a_2 = -8 - (-4) = -4 \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Now, \( a_1 = a = 0, d = -4 \)
\( \therefore \) Next three terms are
\( a_5 = -12 + (-4) = -16 \);
\( a_6 = -16 - 4 = -20 \);
\( a_7 = -20 - 4 = -24 \)
(viii) \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots \)
Here, \( a_2 - a_1 = -\frac{1}{2} - (-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0 \)
and \( a_3 - a_2 = -\frac{1}{2} - (-\frac{1}{2}) = 0 \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Now, \( a_1 = a = -\frac{1}{2}, d = 0 \)
\( \therefore \) Next three terms are \( -\frac{1}{2}, -\frac{1}{2} \) and \( -\frac{1}{2} \)
(ix) \( 1, 3, 9, 27, \dots \)
Here, \( a_2 - a_1 = 3 - 1 = 2 \);
and \( a_3 - a_2 = 9 - 3 = 6 \)
\( a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) The given sequence is not an AP.
(x) \( a, 2a, 3a, 4a, \dots \)
Here, \( a_2 - a_1 = 2a - a = a \)
and \( a_3 - a_2 = 3a - 2a = a \)
\( a_2 - a_1 = a_3 - a_2 \)
\( \therefore \) The given sequence is an AP.
Next three terms are
\( a_5 = 4a + a = 5a \);
\( a_6 = 5a + a = 6a \);
\( a_7 = 6a + a = 7a \)
(xi) \( a, a^2, a^3, a^4, \dots \)
Here, \( a_2 - a_1 = a^2 - a = a(a - 1) \)
and \( a_3 - a_2 = a^3 - a^2 = a^2(a - 1) \)
\( a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) The given sequence is not an AP.
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)
Here, \( a_2 - a_1 = \sqrt{8} - \sqrt{2} = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
and \( a_3 - a_2 = \sqrt{18} - \sqrt{8} = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Next three terms are
\( a_5 = \sqrt{32} + d = \sqrt{16 \times 2} + \sqrt{2} = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50} \)
\( a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72} \)
and \( a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98} \)
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \)
Here, \( a_2 - a_1 = \sqrt{6} - \sqrt{3} \)
and \( a_3 - a_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} \)
\( a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) The given sequence is not an AP.
(xiv) \( 1^2, 3^2, 5^2, 7^2, \dots \)
Here, \( a_2 - a_1 = 9 - 1 = 8 \)
and \( a_3 - a_2 = 25 - 9 = 16 \)
\( a_3 - a_2 \neq a_2 - a_1 \)
\( \therefore \) The given sequence is not an AP.
(xv) \( 1^2, 5^2, 7^2, 73, \dots \)
Here, \( a_2 - a_1 = 25 - 1 = 24 \)
and \( a_3 - a_2 = 49 - 25 = 24 \)
\( a_3 - a_2 = a_2 - a_1 \)
\( \therefore \) The given sequence is an AP.
Now, \( a_1 = a = 1, d = 24 \)
\( \therefore a_5 = 73 + 24 = 97 \);
\( a_6 = 97 + 24 = 121 \);
\( a_7 = 121 + 24 = 145 \)

Question. Fill in the blanks in the following table, given that \( a \) is the first term, \( d \) the common difference and \( a_n \) the \( n \)th term of the AP:
(i) \( a = 7, d = 3, n = 8, a_n = \dots \)
(ii) \( a = -18, n = 10, a_n = 0, d = \dots \)
(iii) \( d = -3, n = 18, a_n = -5, a = \dots \)
(iv) \( a = -18.9, d = 2.5, a_n = 3.6, n = \dots \)
(v) \( a = 3.5, d = 0, n = 105, a_n = \dots \)
Answer: (i) \( a = 7, d = 3, n = 8 \)
\( a_n = a + (n - 1)d \)
\( \implies a_8 = 7 + (8 - 1)3 = 7 + 21 = 28 \)
(ii) \( a = -18, n = 10, a_n = 0 \)
\( a_n = a + (n - 1)d \)
\( \implies 0 = -18 + (10 - 1)d \)
\( \implies 18 = 9d \)
\( \therefore d = 2 \)
(iii) \( d = -3, n = 18, a_n = -5 \)
\( a_n = a + (n - 1)d \)
\( \implies -5 = a + (18 - 1)(-3) \)
\( \implies -5 = a - 51 \)
\( \implies a = 46 \)
(iv) \( a = -18.9, d = 2.5, a_n = 3.6 \)
\( a_n = a + (n - 1)d \)
\( \implies 3.6 = -18.9 + (n - 1) \times 2.5 \)
\( \implies 3.6 + 18.9 = (n - 1) \times 2.5 \)
\( \implies \frac{22.5}{2.5} = n - 1 \)
\( \implies n = 9 + 1 = 10 \)
(v) \( a = 3.5, d = 0, n = 105 \)
\( a_n = a + (n - 1)d \)
\( = 3.5 + (105 - 1)0 = 3.5 + 0 = 3.5 \)

Question. Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, ... , is
(A) 97 (B) 77 (C) -77 (D) -87
(ii) 11th term of the AP: \( -3, -\frac{1}{2}, 2, \dots \) is
(A) 28 (B) 22 (C) -38 (D) \( -48\frac{1}{2} \)
Answer: (i) 10, 7, 4 ...
Here, \( a = 10, d = 7 - 10 = -3 \)
\( \therefore a_{30} = a + (30 - 1)d = a + 29d = 10 + 29(-3) = 10 - 87 = -77 \)
\( \therefore \) Correct option is (C).
(ii) \( -3, -\frac{1}{2}, 2, \dots \) Here,
\( a = -3 \) and \( d = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{5}{2} \)
\( \therefore a_{11} = a + 10d = -3 + 10 \times \frac{5}{2} = -3 + 25 = 22 \)
Hence, correct option is (B).

Question. Which term of the AP: 3, 8, 13, 18, ... , is 78?
Answer: Let \( n \)th term is 78, i.e. \( a_n = 78 \).
Given, 3, 8, 13, 18, .......
Here, \( a = 3, d = 8 - 3 = 5 \)
We have, \( a + (n - 1)d = 78 \)
\( \implies 3 + (n - 1)5 = 78 \)
\( \implies (n - 1)5 = 78 - 3 \)
\( \implies (n - 1)5 = 75 \)
\( \implies n - 1 = \frac{75}{5} = 15 \)
\( \implies n = 15 + 1 = 16 \)
Hence, \( a_{16} = 78 \).

Question. Find the number of terms in each of the following APs:
(i) 7, 13, 19, ... , 205
(ii) \( 18, 15\frac{1}{2}, 13, \dots, -47 \)
Answer: (i) Let there are \( n \) terms in the AP, 7, 13, 19, ..., 205
Here, \( a = 7, d = 13 - 7 = 6, a_n = 205 \)
We have, \( a_n = a + (n - 1) d \)
\( \implies 205 = 7 + (n - 1) 6 \)
\( \implies 205 - 7 = (n - 1) 6 \)
\( \implies \frac{198}{6} = n - 1 \)
\( \implies n = 34 \)
(ii) \( 18, 15\frac{1}{2}, 13, \dots, -47 \)
Here, \( a = 18; d = \frac{31}{2} - \frac{18}{1} = -\frac{5}{2} \)
and \( a_n = -47 \)
Now, \( a_n = a + (n - 1)d \)
\( \implies -47 = 18 + (n - 1)(-\frac{5}{2}) \)
\( \implies -47 - 18 = (n - 1)(-\frac{5}{2}) \)
\( \implies -65 \times (-\frac{2}{5}) = n - 1 \)
\( \implies 26 = n - 1 \)
\( \implies 27 = n \)

Question. Check whether -150 is a term of the AP: 11, 8, 5, 2...
Answer: 11, 8, 5, 2, . . . . . . .
Here, \( a = 11, d = 8 - 11 = -3 \)
Let \( a_n = -150 \)
\( \therefore a + (n - 1) d = -150 \)
\( \implies 11 + (n - 1)(-3) = -150 \)
\( \implies (n - 1)(-3) = -150 - 11 \)
\( \implies -3(n - 1) = -161 \)
\( \implies n - 1 = \frac{-161}{-3} \)
\( \implies n = \frac{161}{3} + 1 = \frac{164}{3} = 54\frac{2}{3} \)
which is not an integral number.
Hence, -150 is not a term of the AP.

Question. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer: Given; \( a_{11} = 38 \) and \( a_{16} = 73 \)
\( \implies a + 10d = 38 \) and \( a + 15d = 73 \)
\( \implies a + 15d - a - 10d = 73 - 38 \)
\( \implies 5d = 35 \)
\( \implies d = \frac{35}{5} = 7 \)
\( \therefore a_{11} = a + 10 \times 7 = 38 \)
\( \implies a = 38 - 70 = -32 \)
\( \therefore a_{31} = a + 30d = -32 + 30 \times 7 = -32 + 210 = 178 \)

Question. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Last term \( = 106 \)
Given, \( a_{50} = 106 \)
\( \implies a + 49d = 106 \dots (i) \)
and \( a_3 = 12 \)
\( \implies a + 2d = 12 \dots (ii) \)
Subtracting (ii) from (i), we have
\( a + 49d - a - 2d = 106 - 12 \)
\( \implies 47d = 94 \)
\( \implies d = \frac{94}{47} = 2 \)
Now, \( a + 2d = 12 \)
\( \implies a + 2 \times 2 = 12 \)
\( \implies a + 4 = 12 \)
\( \implies a = 12 - 4 = 8 \)
\( \therefore a_{29} = a + 28d = 8 + 28 \times 2 = 8 + 56 = 64 \)

Question. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Answer: Given, \( a_3 = 4 \) and \( a_9 = -8 \)
\( \implies a + 2d = 4 \dots (i) \)
and \( a + 8d = -8 \dots (ii) \)
Subtracting (i) from (ii), we have
\( a + 8d - a - 2d = -8 - 4 \)
\( \implies 6d = -12 \)
\( \implies d = \frac{-12}{6} = -2 \)
Now, \( a + 2d = 4 \)
\( \implies a + 2(-2) = 4 \)
\( \implies a - 4 = 4 \)
\( \implies a = 4 + 4 = 8 \)
Let \( a_n = 0 \)
\( \implies a + (n - 1)d = 0 \)
\( \implies 8 + (n - 1)(-2) = 0 \)
\( \implies 8 = 2(n - 1) \)
\( \implies \frac{8}{2} = n - 1 \)
\( \implies 4 = n - 1 \)
\( \implies n = 4 + 1 = 5 \)
Hence, 5th term is zero.

Question. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Answer: Given, \( a_{17} - a_{10} = 7 \)
\( \implies (a + 16d) - (a + 9d) = 7 \)
\( \implies 7d = 7 \)
\( \implies d = \frac{7}{7} = 1 \)

Question. Which term of the AP: 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Answer: 3, 15, 27, 39, ...
Here, \( a = 3, d = 15 - 3 = 12 \)
Let \( a_n = 132 + a_{54} \)
\( \implies a_n - a_{54} = 132 \)
\( \implies (n - 54)12 = 132 \)
[\( \because a_n - a_k = (n - k) d \)]
\( \implies n - 54 = \frac{132}{12} \)
\( \implies n - 54 = 11 \)
\( \implies n = 11 + 54 = 65 \)
\( \therefore a_n = 65 \)

Question. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Answer: Let \( a \) and \( A \) be the first term of two APs and \( d \) be the common difference.
Given, \( a_{100} - A_{100} = 100 \)
\( \implies a + 99d - A - 99d = 100 \)
\( \implies a - A = 100 \)
\( \implies a_{1000} - A_{1000} = a + 999d - A - 999d \)
\( = a - A = 100 \)
\( \implies a_{1000} - A_{1000} = 100 \)

Question. How many three-digit numbers are divisible by 7?
Answer: The three-digit numbers which are divisible by 7 are 105, 112, 119, ........ 994
Here, \( a = 105, d = 7, a_n = 994 \)
\( \therefore a + (n - 1) d = 994 \)
\( \implies 105 + (n - 1) 7 = 994 \)
\( \implies (n - 1) 7 = 994 - 105 \)
\( \implies 7 (n - 1) = 889 \)
\( \implies n - 1 = \frac{889}{7} = 127 \)
\( \implies n = 127 + 1 = 128 \)

Question. How many multiples of 4 lie between 10 and 250?
Answer: The multiples of 4 between 10 and 250 be 12, 16, 20, 24, .... 248
Here, \( a = 12, d = 4, a_n = 248 \)
\( \implies a_n = a + (n - 1) d \)
\( \implies 248 = 12 + (n - 1) 4 \)
\( \implies 248 - 12 = (n - 1) 4 \)
\( \implies \frac{236}{4} = n - 1 \)
\( \implies 59 = n - 1 \)
\( \implies n = 59 + 1 = 60 \)

Question. For what value of \( n \), are the nth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?
Answer: First AP: 63, 65, 67, ...
Here, \( a = 63, d = 65 - 63 = 2 \)
\( a_n = a + (n - 1) d \)
\( = 63 + (n - 1) 2 \)
\( = 63 + 2n - 2 \)
\( \implies a_n = 61 + 2n \)
Second AP: 3, 10, 17 ......
Here, \( a = 3, d = 10 - 3 = 7 \)
\( a_n = a + (n - 1) d \)
\( = 3 + (n - 1) 7 \)
\( = 3 + 7n - 7 = 7n - 4 \)
Now, \( a_n \) of first AP = \( a_n \) of second AP
\( \implies 61 + 2n = 7n - 4 \)
\( \implies 61 + 4 = 7n - 2n \)
\( \implies 65 = 5n \)
\( \implies n = 13 \)

Question. Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.
Answer: Given, \( a_3 = 16 \)
\( \implies a + 2d = 16 \) and \( a_7 - a_5 = 12 \)
\( \implies a + 6d - a - 4d = 12 \)
\( \implies 2d = 12 \implies d = 6 \)
Since \( a + 2d = 16 \)
\( \implies a + 2(6) = 16 \implies a + 12 = 16 \)
\( \implies a = 16 - 12 = 4 \)
\( \therefore \) The required AP is 4, 4 + 6, 10 + 6, 16 + 6 = 4, 10, 16, 22, ...

Question. Find the 20th term from the last term of the AP: 3, 8, 13, ... , 253.
Answer: Given, AP is 3, 8, 13, ....., 253
Here, \( a = 3, d = 8 - 3 = 5 \)
First term from the last = 253 then \( d = - 5 \)
\( a_{20} = a + 19d \)
\( = 253 + 19 (- 5) \)
\( = 253 - 95 = 158 \)

Question. The sum of the 4th and 8th 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: Given, \( a_4 + a_8 = 24 \) and \( a_6 + a_{10} = 44 \)
\( \implies a + 3d + a + 7d = 24 \)
and \( a + 5d + a + 9d = 44 \)
\( \implies 2a + 10d = 24 \) and \( 2a + 14d = 44 \)
\( 2a + 14d - 2a - 10d = 44 - 24 \)
\( \implies 4d = 20 \implies d = \frac{20}{4} = 5 \)
Now, \( 2a + 10d = 24 \)
\( \implies 2a + 10 \times 5 = 24 \)
\( \implies 2a = 24 - 50 \)
\( \implies a = \frac{-26}{2} = -13 \)
Hence the first three terms are:
-13, -13 + 5, -8 + 5, i.e. -13, -8 and -3

Question. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?
Answer: Here, \( a = \text{Rs. } 5000, d = \text{Rs. } 200 \)
Let \( a_n = \text{Rs. } 7000 \)
We have, \( a + (n - 1) d = 7000 \)
\( \implies 5000 + (n - 1) 200 = 7000 \)
\( \implies (n - 1) 200 = 7000 - 5000 \)
\( \implies (n - 1) 200 = 2000 \)
\( \implies (n - 1) = \frac{2000}{200} \)
\( \implies n - 1 = 10 \)
\( \implies n = 11 \)
\( \implies 1995 + 11 = 2006 \)
Hence, in 2006, Subba Rao’s income will reach Rs. 7000.

Question. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find \( n \).
Answer: Given, \( a = \text{Rs. } 5, d = \text{Rs. } 1.75 \)
\( a_n = \text{Rs. } 20.75 \)
\( \implies a + (n - 1) d = 20.75 \)
\( \implies 5 + (n - 1) 1.75 = 20.75 \)
\( \implies (n - 1) \times 1.75 = 20.75 - 5 \)
\( \implies (n - 1) \times 1.75 = 15.75 \)
\( \implies n - 1 = \frac{15.75}{1.75} = 9 \)
\( \implies n = 9 + 1 \)
\( \implies n = 10 \)
Hence, in 10th week, Ramkali’s saving will be Rs. 20.75.