ARITHMETIC PROGRESSIONS - Q&A

1. 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 ₹ 15 for the first km and ₹ 8 for each additional km.

Answer:
Let a_n be the fare for n km.
Fare for 1st km (a₁) = 15
Fare for 2nd km (a₂) = 15 + 8 = 23
Fare for 3rd km (a₃) = 23 + 8 = 31
Fare for 4th km (a₄) = 31 + 8 = 39
List of numbers: 15, 23, 31, 39, ...
Here, difference between consecutive terms = 23 - 15 = 8, 31 - 23 = 8, etc.
Since the common difference is constant (d = 8), it is an AP.

(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.

Answer:
Let the initial volume of air be V.
1st term (a₁) = V
Air removed = (1/4)V. Remaining air (a₂) = V - (1/4)V = (3/4)V
Air removed next = 1/4 of (3/4)V = (3/16)V. Remaining air (a₃) = (3/4)V - (3/16)V = (12V - 3V)/16 = (9/16)V = (3/4)²V
List of numbers: V, (3/4)V, (3/4)²V, ...
Difference a₂ - a₁ = -1/4 V
Difference a₃ - a₂ = 9/16 V - 12/16 V = -3/16 V
Since the difference is not constant, it is not an AP.

(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.

Answer:
Cost for 1st metre = 150
Cost for 2nd metre = 150 + 50 = 200
Cost for 3rd metre = 200 + 50 = 250
List: 150, 200, 250, ...
Difference is constant (50). It is an AP.

(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.

Answer:
Principal (P) = 10000, Rate (R) = 8%
Amount A_n = P(1 + R/100)ⁿ
Year 1: 10000 (1 + 8/100)¹ = 10800
Year 2: 10000 (1 + 8/100)² = 11664
Year 3: 10000 (1 + 8/100)³ ≈ 12597
Difference 2-1 = 864. Difference 3-2 = 933.
Since differences are not same, it is not an AP.

2. 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

Answer:
a₁ = 10
a₂ = 10 + 10 = 20
a₃ = 20 + 10 = 30
a₄ = 30 + 10 = 40
Terms: 10, 20, 30, 40

(ii) a = -2, d = 0

Answer:
a₁ = -2
a₂ = -2 + 0 = -2
a₃ = -2 + 0 = -2
a₄ = -2 + 0 = -2
Terms: -2, -2, -2, -2

(iii) a = 4, d = -3

Answer:
a₁ = 4
a₂ = 4 + (-3) = 1
a₃ = 1 + (-3) = -2
a₄ = -2 + (-3) = -5
Terms: 4, 1, -2, -5

(iv) a = -1, d = 1/2

Answer:
a₁ = -1
a₂ = -1 + 0.5 = -0.5 (or -1/2)
a₃ = -0.5 + 0.5 = 0
a₄ = 0 + 0.5 = 0.5 (or 1/2)
Terms: -1, -1/2, 0, 1/2

(v) a = -1.25, d = -0.25

Answer:
a₁ = -1.25
a₂ = -1.25 - 0.25 = -1.50
a₃ = -1.50 - 0.25 = -1.75
a₄ = -1.75 - 0.25 = -2.00
Terms: -1.25, -1.50, -1.75, -2.00

3. For the following APs, write the first term and the common difference:

(i) 3, 1, -1, -3, ...

Answer:
First term (a) = 3
Common difference (d) = Second term - First term = 1 - 3 = -2

(ii) -5, -1, 3, 7, ...

Answer:
First term (a) = -5
Common difference (d) = -1 - (-5) = -1 + 5 = 4

(iii) 1/3, 5/3, 9/3, 13/3, ...

Answer:
First term (a) = 1/3
Common difference (d) = 5/3 - 1/3 = 4/3

(iv) 0.6, 1.7, 2.8, 3.9, ...

Answer:
First term (a) = 0.6
Common difference (d) = 1.7 - 0.6 = 1.1

4. 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, ...

Answer:
a₂ - a₁ = 4 - 2 = 2
a₃ - a₂ = 8 - 4 = 4
Difference is not constant. Not an AP.

(ii) 2, 5/2, 3, 7/2, ...

Answer:
5/2 - 2 = 0.5
3 - 5/2 = 0.5
Difference is constant (d = 0.5 or 1/2). It is an AP.
Next three terms:
7/2 + 1/2 = 4
4 + 1/2 = 9/2
9/2 + 1/2 = 5
Terms: 4, 9/2, 5

(iii) -1.2, -3.2, -5.2, -7.2, ...

Answer:
-3.2 - (-1.2) = -2.0
-5.2 - (-3.2) = -2.0
Difference is constant (d = -2). It is an AP.
Next three terms:
-7.2 - 2 = -9.2
-9.2 - 2 = -11.2
-11.2 - 2 = -13.2
Terms: -9.2, -11.2, -13.2

(iv) -10, -6, -2, 2, ...

Answer:
-6 - (-10) = 4
-2 - (-6) = 4
Difference is constant (d = 4). It is an AP.
Next three terms:
2 + 4 = 6
6 + 4 = 10
10 + 4 = 14
Terms: 6, 10, 14

(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ...

Answer:
(3 + √2) - 3 = √2
(3 + 2√2) - (3 + √2) = √2
Difference is constant (d = √2). It is an AP.
Next terms:
3 + 4√2
3 + 5√2
3 + 6√2
Terms: 3 + 4√2, 3 + 5√2, 3 + 6√2

(vi) 0.2, 0.22, 0.222, 0.2222, ...

Answer:
0.22 - 0.2 = 0.02
0.222 - 0.22 = 0.002
Difference is not constant. Not an AP.

(vii) 0, -4, -8, -12, ...

Answer:
-4 - 0 = -4
-8 - (-4) = -4
Difference is constant (d = -4). It is an AP.
Next terms: -16, -20, -24.

(viii) -1/2, -1/2, -1/2, -1/2, ...

Answer:
Difference is 0. It is an AP.
Next terms: -1/2, -1/2, -1/2.

(ix) 1, 3, 9, 27, ...

Answer:
3 - 1 = 2
9 - 3 = 6
Difference is not constant. Not an AP.

(x) a, 2a, 3a, 4a, ...

Answer:
2a - a = a
3a - 2a = a
Difference is constant (d = a). It is an AP.
Next terms: 5a, 6a, 7a.

(xi) a, a², a³, a⁴, ...

Answer:
a² - a = a(a-1)
a³ - a² = a²(a-1)
Difference is not constant. Not an AP.

(xii) √2, √8, √18, √32, ...

Answer:
Rewrite: √2, 2√2, 3√2, 4√2
2√2 - √2 = √2
3√2 - 2√2 = √2
Difference is constant (d = √2). It is an AP.
Next terms: 5√2 (√50), 6√2 (√72), 7√2 (√98).

(xiii) √3, √6, √9, √12, ...

Answer:
√6 - √3
√9 - √6 = 3 - √6
Difference is not constant. Not an AP.

(xiv) 1², 3², 5², 7², ...

Answer:
Values: 1, 9, 25, 49
9 - 1 = 8
25 - 9 = 16
Difference is not constant. Not an AP.

(xv) 1², 5², 7², 73, ...

Answer:
Values: 1, 25, 49, 73
25 - 1 = 24
49 - 25 = 24
73 - 49 = 24
Difference is constant (d = 24). It is an AP.
Next terms:
73 + 24 = 97
97 + 24 = 121
121 + 24 = 145
Terms: 97, 121, 145

1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_n the nth term of the AP:

(i) a = 7, d = 3, n = 8, a_n = ...

Answer:
a_n = a + (n - 1)d
a₈ = 7 + (8 - 1)3
a₈ = 7 + 7 × 3 = 7 + 21 = 28

(ii) a = -18, d = ..., n = 10, a_n = 0

Answer:
0 = -18 + (10 - 1)d
18 = 9d
d = 2

(iii) a = ..., d = -3, n = 18, a_n = -5

Answer:
-5 = a + (18 - 1)(-3)
-5 = a + 17(-3)
-5 = a - 51
a = 51 - 5 = 46

(iv) a = -18.9, d = 2.5, n = ..., a_n = 3.6

Answer:
3.6 = -18.9 + (n - 1)(2.5)
3.6 + 18.9 = (n - 1)(2.5)
22.5 = (n - 1)(2.5)
n - 1 = 22.5 / 2.5 = 9
n = 9 + 1 = 10

(v) a = 3.5, d = 0, n = 105, a_n = ...

Answer:
a_n = 3.5 + (105 - 1)(0)
a_n = 3.5 + 0 = 3.5

2. 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

Answer:
a = 10, d = 7 - 10 = -3, n = 30
a₃₀ = a + (30 - 1)d = 10 + 29(-3) = 10 - 87 = -77
Option (C) is correct.

(ii) 11th term of the AP: -3, -1/2, 2, ..., is
(A) 28 (B) 22 (C) -38 (D) -48 1/2

Answer:
a = -3, d = -1/2 - (-3) = -0.5 + 3 = 2.5 (or 5/2), n = 11
a₁₁ = -3 + (11 - 1)(2.5) = -3 + 10(2.5) = -3 + 25 = 22
Option (B) is correct.

3. In the following APs, find the missing terms in the boxes:

(i) 2, [ ], 26

Answer:
Here a = 2, a₃ = 26.
a₃ = a + 2d ⇒ 26 = 2 + 2d ⇒ 24 = 2d ⇒ d = 12.
Missing term a₂ = a + d = 2 + 12 = 14.

(ii) [ ], 13, [ ], 3

Answer:
a₂ = a + d = 13
a₄ = a + 3d = 3
Subtracting (1) from (2): 2d = -10 ⇒ d = -5.
First term a = 13 - d = 13 - (-5) = 18.
Third term a₃ = 13 + d = 13 - 5 = 8.
Box values: 18, 8

(iii) 5, [ ], [ ], 9 1/2

Answer:
a = 5, a₄ = 9.5 (or 19/2).
a₄ = a + 3d ⇒ 9.5 = 5 + 3d ⇒ 4.5 = 3d ⇒ d = 1.5.
Second term = 5 + 1.5 = 6.5 (or 6 1/2).
Third term = 6.5 + 1.5 = 8.
Box values: 6 1/2, 8

(iv) -4, [ ], [ ], [ ], [ ], 6

Answer:
a = -4, a₆ = 6.
a + 5d = 6 ⇒ -4 + 5d = 6 ⇒ 5d = 10 ⇒ d = 2.
Terms: -4 + 2 = -2, -2 + 2 = 0, 0 + 2 = 2, 2 + 2 = 4.
Box values: -2, 0, 2, 4

(v) [ ], 38, [ ], [ ], [ ], -22

Answer:
a₂ = 38, a₆ = -22.
a + d = 38 ... (1)
a + 5d = -22 ... (2)
(2) - (1) ⇒ 4d = -60 ⇒ d = -15.
From (1): a - 15 = 38 ⇒ a = 53.
Term 3: 38 - 15 = 23.
Term 4: 23 - 15 = 8.
Term 5: 8 - 15 = -7.
Box values: 53, 23, 8, -7

4. Which term of the AP: 3, 8, 13, 18, ..., is 78?

Answer:
a = 3, d = 5, a_n = 78.
78 = 3 + (n - 1)5
75 = 5(n - 1)
15 = n - 1 ⇒ n = 16.
16th term

5. Find the number of terms in each of the following APs:

(i) 7, 13, 19, ..., 205

Answer:
a = 7, d = 6, a_n = 205.
205 = 7 + (n - 1)6
198 = 6(n - 1)
33 = n - 1 ⇒ n = 34.
34 terms

(ii) 18, 15 1/2, 13, ..., -47

Answer:
a = 18, d = 15.5 - 18 = -2.5, a_n = -47.
-47 = 18 + (n - 1)(-2.5)
-65 = (n - 1)(-2.5)
n - 1 = 26 ⇒ n = 27.
27 terms

6. Check whether -150 is a term of the AP: 11, 8, 5, 2...

Answer:
a = 11, d = -3.
Let -150 be the nth term.
-150 = 11 + (n - 1)(-3)
-161 = (n - 1)(-3)
n - 1 = 161/3
n = 161/3 + 1 = 164/3 ≈ 54.66
Since n is not an integer, -150 is not a term of this AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Answer:
a₁₁ = a + 10d = 38
a₁₆ = a + 15d = 73
Subtracting: 5d = 35 ⇒ d = 7.
a + 70 = 38 ⇒ a = 38 - 70 = -32.
a₃₁ = a + 30d = -32 + 30(7) = -32 + 210 = 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

Answer:
n = 50, a₃ = 12, a₅₀ = 106.
a + 2d = 12
a + 49d = 106
Subtracting: 47d = 94 ⇒ d = 2.
a + 2(2) = 12 ⇒ a = 8.
a₂₉ = 8 + 28(2) = 8 + 56 = 64.

9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Answer:
a + 2d = 4
a + 8d = -8
Subtracting: 6d = -12 ⇒ d = -2.
a + 2(-2) = 4 ⇒ a - 4 = 4 ⇒ a = 8.
Find n for a_n = 0:
0 = 8 + (n - 1)(-2)
-8 = -2(n - 1)
4 = n - 1 ⇒ n = 5.
The 5th term is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Answer:
a₁₇ - a₁₀ = 7
(a + 16d) - (a + 9d) = 7
7d = 7 ⇒ d = 1.
Common difference is 1.

11. Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?

Answer:
a = 3, d = 12.
a₅₄ = 3 + 53(12) = 3 + 636 = 639.
Required value = 639 + 132 = 771.
771 = 3 + (n - 1)12
768 = 12(n - 1)
64 = n - 1 ⇒ n = 65.
65th term.

12. 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 first AP be a, a+d... and second be b, b+d...
Difference between 100th terms: (a + 99d) - (b + 99d) = a - b = 100.
Difference between 1000th terms: (a + 999d) - (b + 999d) = a - b.
Since a - b = 100, the difference remains 100.

13. How many three-digit numbers are divisible by 7?

Answer:
First 3-digit number divisible by 7 is 105.
Last 3-digit number divisible by 7 is 994.
AP: 105, 112, ..., 994. a = 105, d = 7.
994 = 105 + (n - 1)7
889 = 7(n - 1)
127 = n - 1 ⇒ n = 128.
128 numbers.

14. How many multiples of 4 lie between 10 and 250?

Answer:
First multiple after 10 is 12.
Last multiple before 250 is 248.
AP: 12, 16, ..., 248. a = 12, d = 4.
248 = 12 + (n - 1)4
236 = 4(n - 1)
59 = n - 1 ⇒ n = 60.
60 multiples.

15. For what value of n, are the nth terms of two APs: 63, 65, 67, ... and 3, 10, 17, ... equal?

Answer:
AP1: 63, 65... a=63, d=2. nth term = 63 + 2(n-1) = 61 + 2n.
AP2: 3, 10... a=3, d=7. nth term = 3 + 7(n-1) = 7n - 4.
Equating them: 61 + 2n = 7n - 4
65 = 5n ⇒ n = 13.
13th term.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer:
a₃ = 16 ⇒ a + 2d = 16.
a₇ - a₅ = 12 ⇒ (a + 6d) - (a + 4d) = 12 ⇒ 2d = 12 ⇒ d = 6.
a + 2(6) = 16 ⇒ a + 12 = 16 ⇒ a = 4.
AP: 4, 10, 16, 22...

17. Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.

Answer:
Reverse the AP: 253, 248, ..., 3.
a = 253, d = -5.
20th term = a + 19d = 253 + 19(-5) = 253 - 95 = 158.

18. 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:
(a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ... (1)
(a + 5d) + (a + 9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 ... (2)
(2) - (1): 2d = 10 ⇒ d = 5.
a + 25 = 12 ⇒ a = -13.
Terms: -13, -8, -3

19. Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Answer:
a = 5000, d = 200, a_n = 7000.
7000 = 5000 + (n - 1)200
2000 = 200(n - 1)
10 = n - 1 ⇒ n = 11.
11th year from 1995 is 2005.

20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.

Answer:
a = 5, d = 1.75, a_n = 20.75.
20.75 = 5 + (n - 1)1.75
15.75 = (n - 1)1.75
n - 1 = 15.75 / 1.75 = 1575 / 175 = 9.
n = 9 + 1 = 10.

1. Find the sum of the following APs:

(i) 2, 7, 12, ..., to 10 terms.

Answer:
a = 2, d = 5, n = 10.
S₁₀ = (10/2) [2(2) + (9)5] = 5 [4 + 45] = 5 × 49 = 245.

(ii) -37, -33, -29, ..., to 12 terms.

Answer:
a = -37, d = 4, n = 12.
S₁₂ = (12/2) [2(-37) + 11(4)] = 6 [-74 + 44] = 6 × (-30) = -180.

(iii) 0.6, 1.7, 2.8, ..., to 100 terms.

Answer:
a = 0.6, d = 1.1, n = 100.
S₁₀₀ = (100/2) [2(0.6) + 99(1.1)] = 50 [1.2 + 108.9] = 50 × 110.1 = 5505.

(iv) 1/15, 1/12, 1/10, ..., to 11 terms.

Answer:
a = 1/15, d = 1/12 - 1/15 = (5-4)/60 = 1/60, n = 11.
S₁₁ = (11/2) [2(1/15) + 10(1/60)]
= (11/2) [2/15 + 1/6]
= (11/2) [(4 + 5)/30] = (11/2) [9/30] = 99/60 = 33/20.

2. Find the sums given below:

(i) 7 + 10 1/2 + 14 + ... + 84

Answer:
a = 7, d = 3.5, l (last term) = 84.
Find n: 84 = 7 + (n - 1)3.5 ⇒ 77 = 3.5(n - 1) ⇒ 22 = n - 1 ⇒ n = 23.
Sum = (n/2)(a + l) = (23/2)(7 + 84) = (23/2)(91) = 2093/2 = 1046 1/2 (or 1046.5).

(ii) 34 + 32 + 30 + ... + 10

Answer:
a = 34, d = -2, l = 10.
10 = 34 + (n - 1)(-2) ⇒ -24 = -2(n - 1) ⇒ 12 = n - 1 ⇒ n = 13.
Sum = (13/2)(34 + 10) = (13/2)(44) = 13 × 22 = 286.

(iii) -5 + (-8) + (-11) + ... + (-230)

Answer:
a = -5, d = -3, l = -230.
-230 = -5 + (n - 1)(-3) ⇒ -225 = -3(n - 1) ⇒ 75 = n - 1 ⇒ n = 76.
Sum = (76/2)(-5 - 230) = 38 × (-235) = -8930.

3. In an AP:

(i) given a = 5, d = 3, a_n = 50, find n and S_n.

Answer:
50 = 5 + (n - 1)3 ⇒ 45 = 3(n - 1) ⇒ 15 = n - 1 ⇒ n = 16.
S₁₆ = (16/2)(5 + 50) = 8 × 55 = 440.

(ii) given a = 7, a₁₃ = 35, find d and S₁₃.

Answer:
35 = 7 + 12d ⇒ 28 = 12d ⇒ d = 28/12 = 7/3.
S₁₃ = (13/2)(7 + 35) = (13/2)(42) = 13 × 21 = 273.

(iii) given a₁₂ = 37, d = 3, find a and S₁₂.

Answer:
37 = a + 11(3) ⇒ 37 = a + 33 ⇒ a = 4.
S₁₂ = (12/2)(4 + 37) = 6 × 41 = 246.

(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.

Answer:
a + 2d = 15 ... (1)
S₁₀ = (10/2)[2a + 9d] = 125 ⇒ 5[2a + 9d] = 125 ⇒ 2a + 9d = 25 ... (2)
From (1), 2a + 4d = 30. Subtract from (2): 5d = -5 ⇒ d = -1.
a + 2(-1) = 15 ⇒ a = 17.
a₁₀ = a + 9d = 17 - 9 = 8.

(v) given d = 5, S₉ = 75, find a and a₉.

Answer:
75 = (9/2)[2a + 8(5)] ⇒ 150/9 = 2a + 40 ⇒ 50/3 - 40 = 2a ⇒ (50-120)/3 = 2a ⇒ -70/3 = 2a.
a = -35/3.
a₉ = a + 8d = -35/3 + 40 = (-35 + 120)/3 = 85/3.

(vi) given a = 2, d = 8, S_n = 90, find n and a_n.

Answer:
90 = (n/2)[4 + (n - 1)8] ⇒ 180 = n[4 + 8n - 8] ⇒ 180 = 4n² - 4n.
4n² - 4n - 180 = 0 ⇒ n² - n - 45 = 0.
(n - 6)(n + 5) = 0. Since n > 0, n = 6.
a₆ = 2 + 5(8) = 42.

(vii) given a = 8, a_n = 62, S_n = 210, find n and d.

Answer:
210 = (n/2)(8 + 62) ⇒ 420 = 70n ⇒ n = 6.
62 = 8 + 5d ⇒ 54 = 5d ⇒ d = 54/5.

(viii) given a_n = 4, d = 2, S_n = -14, find n and a.

Answer:
4 = a + (n - 1)2 ⇒ a = 4 - 2n + 2 = 6 - 2n.
-14 = (n/2)(a + 4). Substitute a: -28 = n(6 - 2n + 4) ⇒ -28 = n(10 - 2n).
-28 = 10n - 2n² ⇒ 2n² - 10n - 28 = 0 ⇒ n² - 5n - 14 = 0.
(n - 7)(n + 2) = 0. n = 7.
a = 6 - 2(7) = -8.

(ix) given a = 3, n = 8, S = 192, find d.

Answer:
192 = (8/2)[6 + 7d] ⇒ 192 = 4[6 + 7d] ⇒ 48 = 6 + 7d ⇒ 42 = 7d ⇒ d = 6.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Answer:
144 = (9/2)(a + 28) ⇒ 288/9 = a + 28 ⇒ 32 = a + 28 ⇒ a = 4.

4. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?

Answer:
a = 9, d = 8, S_n = 636.
636 = (n/2)[18 + (n - 1)8] ⇒ 1272 = n[18 + 8n - 8] ⇒ 1272 = 10n + 8n².
8n² + 10n - 1272 = 0 ⇒ 4n² + 5n - 636 = 0.
Using quadratic formula: n = [-5 ± √(25 - 4(4)(-636))] / 8
= [-5 ± √(25 + 10176)] / 8 = [-5 ± √10201] / 8 = (-5 ± 101) / 8.
n = 96 / 8 = 12. (Ignore negative).

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer:
a = 5, l = 45, S = 400.
400 = (n/2)(5 + 45) ⇒ 800 = 50n ⇒ n = 16.
45 = 5 + 15d ⇒ 40 = 15d ⇒ d = 40/15 = 8/3.

6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer:
a = 17, l = 350, d = 9.
350 = 17 + (n - 1)9 ⇒ 333 = 9(n - 1) ⇒ 37 = n - 1 ⇒ n = 38.
Sum = (38/2)(17 + 350) = 19 × 367 = 6973.

7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Answer:
149 = a + 21(7) ⇒ 149 = a + 147 ⇒ a = 2.
Sum = (22/2)(2 + 149) = 11 × 151 = 1661.

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Answer:
d = 18 - 14 = 4.
a = 14 - 4 = 10.
S₅₁ = (51/2)[2(10) + 50(4)] = (51/2)[20 + 200] = (51/2)[220] = 51 × 110 = 5610.

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Answer:
S₇ = 49 ⇒ (7/2)[2a + 6d] = 49 ⇒ a + 3d = 7 ... (1)
S₁₇ = 289 ⇒ (17/2)[2a + 16d] = 289 ⇒ a + 8d = 17 ... (2)
(2) - (1): 5d = 10 ⇒ d = 2. a = 1.
S_n = (n/2)[2(1) + (n-1)2] = (n/2)[2 + 2n - 2] = (n/2)[2n] = n².

10. Show that a₁, a₂, ..., a_n, ... form an AP where a_n is defined as below:

(i) a_n = 3 + 4n

Answer:
a₁ = 7, a₂ = 11, a₃ = 15. Difference is constant (4). It is an AP.
S₁₅ = (15/2)(7 + a₁₅). a₁₅ = 3 + 60 = 63.
S₁₅ = (15/2)(7 + 63) = (15/2)(70) = 15 × 35 = 525.

(ii) a_n = 9 - 5n

Answer:
a₁ = 4, a₂ = -1, a₃ = -6. Difference is constant (-5). It is an AP.
S₁₅ = (15/2)(4 + a₁₅). a₁₅ = 9 - 75 = -66.
S₁₅ = (15/2)(4 - 66) = (15/2)(-62) = 15 × (-31) = -465.

11. If the sum of the first n terms of an AP is 4n - n², what is the first term (that is S₁)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Answer:
S_n = 4n - n².
First term (a₁) = S₁ = 4(1) - 1 = 3.
Sum of first two terms (S₂) = 4(2) - 4 = 4.
Second term (a₂) = S₂ - S₁ = 4 - 3 = 1.
S₃ = 4(3) - 9 = 3. a₃ = S₃ - S₂ = 3 - 4 = -1.
S₉ = 36 - 81 = -45. S₁₀ = 40 - 100 = -60.
a₁₀ = S₁₀ - S₉ = -60 - (-45) = -15.
a_n = S_n - S_n-1 = (4n - n²) - [4(n - 1) - (n - 1)²]
= 4n - n² - [4n - 4 - (n² - 2n + 1)]
= 4n - n² - [4n - 4 - n² + 2n - 1]
= 4n - n² - 6n + n² + 5
= 5 - 2n.

12. Find the sum of the first 40 positive integers divisible by 6.

Answer:
6, 12, 18, ... a = 6, d = 6, n = 40.
S₄₀ = (40/2)[12 + 39(6)] = 20[12 + 234] = 20[246] = 4920.

13. Find the sum of the first 15 multiples of 8.

Answer:
8, 16, 24, ... a = 8, d = 8, n = 15.
S₁₅ = (15/2)[16 + 14(8)] = (15/2)[16 + 112] = (15/2)[128] = 15 × 64 = 960.

14. Find the sum of the odd numbers between 0 and 50.

Answer:
1, 3, 5, ..., 49. a = 1, l = 49.
49 = 1 + (n - 1)2 ⇒ 48 = 2(n - 1) ⇒ n = 25.
Sum = (25/2)(1 + 49) = (25/2)(50) = 25 × 25 = 625.

15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Answer:
a = 200, d = 50, n = 30.
S₃₀ = (30/2)[2(200) + 29(50)] = 15[400 + 1450] = 15[1850] = 27750.
Penalty = ₹ 27750.

16. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.

Answer:
S₇ = 700, n = 7, d = -20.
700 = (7/2)[2a + 6(-20)] ⇒ 200 = 2a - 120 ⇒ 320 = 2a ⇒ a = 160.
Prizes: ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40.

17. In a school, students thought of planting trees... (according to class number, 3 sections each). How many trees will be planted by the students?

Answer:
Class I: 3 sections × 1 tree = 3.
Class II: 3 × 2 = 6.
Class III: 3 × 3 = 9.
... Class XII: 3 × 12 = 36.
AP: 3, 6, 9, ..., 36. n = 12.
Sum = (12/2)(3 + 36) = 6 × 39 = 234 trees.

18. A spiral is made up of successive semicircles... What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Answer:
Length of semicircle = πr.
l₁ = π(0.5), l₂ = π(1.0), l₃ = π(1.5)...
Total length = π(0.5) + π(1.0) + ... (13 terms)
= π(0.5)[1 + 2 + ... + 13]
= (22/7)(0.5) × (13/2)(1 + 13) (Sum of n integers)
= (11/7) × (13 × 14)/2 = (11/7) × 91 = 11 × 13 = 143 cm.

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Answer:
a = 20, d = -1, S_n = 200.
200 = (n/2)[40 + (n - 1)(-1)] ⇒ 400 = n[40 - n + 1] ⇒ 400 = 41n - n².
n² - 41n + 400 = 0 ⇒ (n - 16)(n - 25) = 0.
If n = 25, a₂₅ = 20 - 24 = -4 (Negative logs impossible).
So, n = 16 rows.
Top row (a₁₆) = 20 - 15 = 5 logs.

20. In a potato race... What is the total distance the competitor has to run?

Answer:
Distances run: 2×5, 2×(5+3), 2×(5+6)...
10, 16, 22, ... (10 terms).
a = 10, d = 6, n = 10.
Sum = (10/2)[20 + 9(6)] = 5[20 + 54] = 5[74] = 370 m.

1. Which term of the AP: 121, 117, 113, ..., is its first negative term?

Answer:
a = 121, d = -4.
a_n < 0 ⇒ 121 + (n - 1)(-4) < 0
121 - 4n + 4 < 0
125 < 4n ⇒ n > 31.25.
First integer n is 32. 32nd term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Answer:
a₃ + a₇ = 6 ⇒ (a + 2d) + (a + 6d) = 6 ⇒ 2a + 8d = 6 ⇒ a + 4d = 3 ⇒ a = 3 - 4d.
(a + 2d)(a + 6d) = 8.
Substitute a: (3 - 4d + 2d)(3 - 4d + 6d) = 8 ⇒ (3 - 2d)(3 + 2d) = 8.
9 - 4d² = 8 ⇒ 4d² = 1 ⇒ d = ±1/2.
Case 1: d = 1/2, a = 1. S₁₆ = (16/2)[2(1) + 15(0.5)] = 8[2 + 7.5] = 76.
Case 2: d = -1/2, a = 5. S₁₆ = (16/2)[10 - 7.5] = 8[2.5] = 20.
Sum is 76 or 20.

3. A ladder has rungs 25 cm apart... (Total length of wood?)

Answer:
Top rung = 25 cm, Bottom rung = 45 cm.
Total height = 2.5 m = 250 cm.
Number of rungs = (250/25) + 1 = 11.
Sum = (11/2)(25 + 45) = 5.5 × 70 = 385 cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Answer:
Sum(1 to x-1) = Sum(x+1 to 49).
S_x-1 = S₄₉ - S_x.
[(x-1)x]/2 = [49(50)]/2 - [x(x+1)]/2.
x² - x = 2450 - (x² + x)
2x² = 2450 ⇒ x² = 1225.
x = 35.
House number 35.

5. A small terrace... Calculate the total volume of concrete required to build the terrace.

Answer:
Step 1 volume: 50 × 1/4 × 1/2 = 25/4 m³.
Step 2 volume: 50 × 1/2 × 1/2 = 50/4 m³.
Step 3 volume: 50 × 3/4 × 1/2 = 75/4 m³.
AP with a = 25/4, d = 25/4, n = 15.
Total Volume = (15/2)[2(25/4) + 14(25/4)] = (15/2)[16 × 25/4] = (15/2)[100] = 750 m³.