Showing posts with label (022) NCERT-10-5-Arithmetic Progressions. Show all posts
Showing posts with label (022) NCERT-10-5-Arithmetic Progressions. Show all posts

Sunday, November 5, 2023

163-NCERT-10-5-Arithmetic Progressions - Ex-5.4

NCERT
10th Mathematics
Exercise 5.4
Topic: 5 Arithmetic Progressions

Click here for ⇨ NCERT-10-5-Arithmetic Progressions - Ex- 5.3

EXERCISE 5.4

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?
[Hint: Find n for an < 0]

Solution:

1) According to the problem, a1 = 121, a2 = 117, a3 = 113. . . so d = - 4,
2) We will have to find n using the above information.
an = a + (n – 1) d
an = 121 + (n – 1) (- 4)
an = 121 + (- 4n + 4)
an = 121 - 4n + 4
an = 125 - 4n
3) We have to find which first term is negative. 
an < 0
125 - 4n < 0
125 < 4n
4n > 125
n > 125/4
n > 31.25
4) So the first negative term of this AP is the 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.

Solution:

1) Let the first term be "a" and the common difference be "d".
2) Here a3 + a7 = 6 ---------- equation 1
3) Here a3 x a7 = 8 ---------- equation 2
4) Now we will find a3 and a7,
a) First we will find a3 
an = a + (n – 1) d
a3 = a + (3 – 1) d
a3 = a + 2d ---------- equation 3
b) Now we will find a7 
an = a + (n – 1) d
a7 = a + (7 – 1) d
a7 = a + 6d ---------- equation 4
5) From equations 1, 3, and 4, we have,
a3 + a7 = 6
a + 2d + a + 6d = 6
2a + 8d = 6
2(a + 4d) = 6
(a + 4d) = 6/2
(a + 4d) = 3
a = 3 - 4d ---------- equation 5
6) From equations 2, 3, 4, and 5, we have,
a3 x a7 = 8
(a + 2d) x (a + 6d) = 8
(3 - 4d + 2d) x (3 - 4d + 6d) = 8
(3 - 2d) x (3 + 2d) = 8
(32 - 4d2) = 8
(9 - 4d2) = 8
4d2 = 9 - 8
4d2 = 1
d2 = 1/4
d2 = 1/4
d = ± 1/2
d = 1/2 or - 1/2 ---------- equation 6
7) Put d = 1/2 and d = - 1/2 from equation 6 in equation 5,
a) First we take d = 1/2, we get
a = 3 - 4d
a = 3 - 4(1/2)
a = 3 - 2
a = 1 ---------- equation 7
b) First we take d = - 1/2, we get
a = 3 - 4d
a = 3 - 4(- 1/2)
a = 3 - (- 2)
a = 3 + 2
a = 5 ---------- equation 8
8) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1) d]
a) First we will find sum of first 16 terms with a = 1 and d = 1/2:
Sn = (n/2)[2a + (n - 1) d]
S16 = (16/2)[2(1) + (16 - 1) (1/2)]
S16 = 8[2 + (15) (1/2)]
S16 = 8[2 + (15/2)] 
S16 = 8[(4 + 15)]/2
S16 = 4(19)
S16 = 76 ---------- equation 9
b) First we will find sum of first 16 terms with a = 5 and d = (- 1/2):
Sn = (n/2)[2a + (n - 1) d]
S16 = (16/2)[2(5) + (16 - 1) (- 1/2)]
S16 = 8[10 + (15) (- 1/2)]
S16 = 8[10 - (15/2)] 
S16 = 8[(20 - 15)]/2
S16 = 4(5)
S16 = 20 ---------- equation 10
 9) From equations 9 and 10, we have,
a) S16 = 76 when a = 1 and d = 1/2
b) S16 = 20 when a = 5 and d = - 1/2.

3. A ladder has rungs 25 cm apart. (see the following fig.). The rungs decrease
uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = (250/25) + 1]

Solution:

1) The distance between the rungs is 25 cm.
2) Distance between the top rung and the bottom
     rung is 2½ m. i.e. 5/2 m = 2.5 m = 250 cm.
3) So total number of rungs = [(250/25) + 1] = 11.
4) The length of rungs is decreasing uniformly from bottom to top, so they are in AP.
5) Here, a1 = 45, l = 25 and n = 11, so,
        Sn = (n/2)[a + l]
        S11 = (11/2)[45 + 25]
        S11 = (11/2)[70]
        S11 = 11(35)
        S11 = 385
6) The length of the wood required for the rungs is         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. [Hint : sx - 1 = S49 – Sx]

Solution:

1) Row houses are numbered 1, 2, 3, . . 49.
2) So, a1 = 1, a2 = 2, a3 = 3. . . so d = 1,
3) The sum of the number of houses preceding the xth house will be S(x-1).
Sn = (n/2)[2a + (n – 1) d]
S(x-1) = ((x-1)/2)[2(1) + (x - 1 - 1) (1)]
S(x-1) = ((x-1)/2)[2 + (x - 2)] 
S(x-1) = ((x-1)/2)[x]
S(x-1) = x(x-1)/2 ---------- equation 1
4) Now we will find S49
Sn = (n/2)[2a + (n – 1) d]
S49 = (49/2)[2(1) + (49 - 1) (1)]
S49 = (49/2)[2 + 48] 
S49 = (49/2)(2)[1 + 24]
S49 = (49)[25]
S49 = 1225 ---------- equation 2
5) Now we will find Sx
Sn = (n/2)[2a + (n – 1) d]
Sx = (x/2)[2(1) + (x - 1) (1)]
Sx = (x/2)[2 + (x - 1)] 
Sx = (x/2)[x + 1]
Sx = x(x + 1)/2 ---------- equation 3
6) Now we will find S49 – Sx using equations 2 and 3.
S49 - Sx = 1225 - [x(x + 1)/2] ---------- equation 4
7) According to the problem, S(x-1) =  S49 - Sx  ---------- equation 5
8) From equations 1, 4, and 5, we have,
S(x-1) =  S49 - Sx
x(x-1)/2 =  1225 - [x(x + 1)/2]
x(x-1)/2 =  [2450 - (x(x + 1))]/2
x(x-1) =  [2450 - (x(x + 1))]
x2 - x = 2450 - x2 - x
2x2 = 2450
x2 = 1225
x = ± 35
x = 35 or - 35
9) As the number of houses can't be negative, the value of x will be 35.
 
5. A small terrace at a football ground comprises of 15 steps each of which is
50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see the following fig.). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = (1/4) x (1/2) x 50 m3]
 

1) According to the problem and the figure, we have,
i) The heights of the steps are given bellow:
a) The height of the first step is (1/4) m
b) The height of the second step is (1/4) + (1/4) = (1/2) m
c) The height of the 3rd step is (1/2) + (1/4) = (3/4) m
d) The height of the 4th step is (3/4) + (1/4) = (1) m
2) If we consider the height in the changing form, the width will be the same for all
the steps. i.e. the width will be 1/2 m for all the steps.
3) Here the length of the terrace is 50 m.
4) Here we can find:
The volume of the steps = volume of the cuboid
The volume of the steps = Length x Breadth x Height
5) Now we will find the volumes of the steps:
a) The volume of the first step
= (1/4) x (1/2) x (50)
= (1/8) x (50)
= (50/8)
= (25/4) ---------- equation 1
b) The volume of the second step
= (1/2) x (1/2) x (50)
= (1/4) x (50)
= (50/4) ---------- equation 2 
c) The volume of the 3rd step
= (3/4) x (1/2) x (50)
= (3/4) x (25)
= (75/4) ---------- equation 3 
d) The volume of the 4th step is (3/4) + (1/4) = (1) m 
= (1) x (1/2) x (50)
= (1) x (25)
= (25) ---------- equation 4
6) From equations 1, 2, 3, and 4, volumes of the steps are in AP.
7) So, here, a = 25/4, d = 50/4 - 25/4 = 25/4, and n = 15,
8) so the sum of these 15 steps will be,
Sn = (n/2)[2a + (n – 1) d]
S15 = (15/2)[2(25/4) + (15 – 1) (25/4)]
S15 = (15/2)[2(25/4) + (14) (25/4)]
S15 = (15/2) x (2(25/4)) x [1 + (7)]
S15 = (15/2) x (2(25/4)) x [8]
S15 = [(15 x 2 x 25)/8] x [8]
S15 = (15 x 2 x 25)
S15 = (30 x 25)
S15 = (750)
9) The concrete required to build the terrace is 750 m3.

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Thursday, October 12, 2023

162-NCERT-10-5-Arithmetic Progressions - Ex-5.3

NCERT
10th Mathematics
Exercise 5.3
Topic: 5 Arithmetic Progressions

Click here for ⇨ NCERT-10-5-Arithmetic Progressions - Ex- 5.2

EXERCISE 5.3

Q1. Find the sum of the following APs:
(i) 2, 7, 12, . . ., to 10 terms.     (ii) –37, –33, –29, . . ., to 12 terms.
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.     (iv) 1/15, 1/12, 1/10,  . . ., to 11 terms.

Explanation:

1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) The sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n – 1) d]
3) If an = l is the last term of an AP, then
S = (n/2)[2a + (n – 1) d]
S = (n/2)[a + (a + (n – 1) d)]
S = (n/2)[a + an]
S = (n/2)[a + l].

Solution:

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

1) Here, a = 2, d = 7 - 2 = 5, and n = 10
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n – 1) d]
S = (10/2)[2(2) + (10 – 1) (5)]
S = (5)[4 + 5(9)]
S = (5)[4 + 45]
S = (5)(49)
S = 245
3) So, the sum of the given AP is 245. 
 
(ii) –37, –33, –29, . . ., to 12 terms.

1) Here, a = - 37, d = - 33 + 37 = 4, and n = 12
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n – 1) d]
S = (12/2)[2(- 37) + (12 – 1) (4)]
S = (6)[- 74 + 4(11)]
S = (6)[- 74 + 44]
S = (6)(- 30)
S = - 180
3) So, the sum of the given AP is - 180. 

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

1) Here, a = 0.6, d = 1.7 - 0.6 = 1.1, and n = 100
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n – 1) d]
S = (100/2)[2(0.6) + (100 – 1) (1.1)]
S = (50)[1.2 + 1.1(99)]
S = (50)[1.2 + 108.9]
S = (50)(110.1)
S = 5505
3) So, the sum of the given AP is 5505. 

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

1) Here, a = 1/15, 
d = 1/12 - 1/15
d = (15 - 12)/(12x15)
d = (3)/(12x15)
d = 1/(4x15)
d = 1/60 and
 
n = 11
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n – 1) d]
S = (11/2)[2(1/15) + (11 – 1) (1/60)]
S = (11/2)[2/15 + 10/60]
S = (11/2)[2/15 + 1/6]
S = (11/2)/[(4/30) + (5/30)]
S = (11/2)/(4 + 5)/30 
S = (11/2)/(9/30)
S = (11/2)/(3/10)
S = 33/20 
3) So, the sum of the given AP is 33/20.

Q2. Find the sums given below :
(i) 7 + 10½ + 14 + . . . + 84
(ii) 34 + 32 + 30 + . . . + 10
(iii) –5 + (–8) + (–11) + . . . + (–230)

Solution:

(i) 7 + 10½ + 14 + . . . + 84

1) Here, a = 7, d = 10½ - 7 = 3½ = 7/2, nth term is an = l = 84
2) We know that,
an = a + (n – 1) d
84 = 7 + (n – 1) (7/2)
(n – 1) (7/2) = 84 - 7
(n – 1) (7/2) = 77
(n – 1) = 77(2/7)
(n – 1) = 11(2)
(n – 1) = 22
n = 22 + 1
n = 23 
3) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[a + l]
S = (23/2)[7 + 84]
S = (23/2)[91]
S = 2093/2
S = 1046.5
4) So, the sum of the given AP is 1046.5.

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

1) Here, a = 34, d = 32 - 34 = - 2, nth term is an = 10
2) We know that,
an = a + (n – 1) d
10 = 34 + (n – 1) (- 2)
(n – 1) (- 2) = 10 - 34
(n – 1) (- 2) = - 24
(n – 1) = (- 24)/(- 2)
(n – 1) = 12
n = 12 + 1
n = 13 
3) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[a + l]
S = (13/2)[34 + 10]
S = (13/2)[44]
S = (13)[22]
S = 286 
4) So, the sum of the given AP is 286.

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

1) Here, a = - 5, d = (- 8) - (- 5) = - 3, nth term is an = - 230
2) We know that,
an = a + (n – 1) d
- 230 = - 5 + (n – 1) (- 3)
(n – 1) (- 3) = 5 - 230
(n – 1) (- 3) = - 225
(n – 1) = (- 225)/(- 3)
(n – 1) = 75
n = 75 + 1
n = 76
3) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[a + l]
S = (76/2)[- 5 + (- 230)]
S = (38)[- 235]
S = - 8930 
4) So, the sum of the given AP is - 8930.

Q3. In an AP:
(i) given a = 5, d = 3, an = 50, find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, Sn = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2, Sn = – 14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Solution:

(i) given a = 5, d = 3, an = 50, find n and Sn.

1) Here, a = 5, d = 3, nth term is an = 50.
2) We know that,
an = a + (n – 1) d
50 = 5 + (n – 1) (3)
(n – 1) (3) = 50 - 5
(n – 1) (3) = 45
(n – 1) = 45/3
(n – 1) = 15
n = 15 + 1
n = 16
3) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[a + l]
S = (16/2)[5 + 50]
S = (8)[55]
S = 440 
4) So, Sn = 440, n = 16.

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

1) Here, a = 7, d = ?, 13th term is a13 = 35. Find d and S13
2) We know that,
an = a + (n – 1) d
a13 = 7 + (13 – 1) (d)
a13 = 7 + 12d
35 = 7 + 12d
12d = 35 - 7
12d = 28
d = 28/12 
d = 7/3 --------- equation 1
3) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[a + l]
S13 = (13/2)[7 + 35]
S13 = (13/2)[42]
S13 = (13)[21]
S13 = 273
4) So, S13 = 273, d = 7/3.

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

1) Here, a = ?, d = 3, 12th term is a12 = 37. Find a and S12
2) We know that,
an = a + (n – 1) d
a12 = a + (12 – 1) (3)
37 = a + 3(11)
37 = a + 33
a = 37 - 33
a = 4 --------- equation 1
3) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[a + l]
S12 = (12/2)[4 + 37]
S12 = (6)[41]
S12 = 246 
4) So, S12 = 246, a = 4.  
 
(iv) given a3 = 15, S10 = 125, find d and a10.

1) Here, a3 = 15, S10 = 125. Find d and a10.
2) We know that,
an = a + (n – 1) d
a3 = a + (3 – 1) (d)
15 = a + 2d
a + 2d = 15 --------- equation 1 
3) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1) d]
S10 = (10/2)[2(a) + (10 - 1) d]
125 = (5)[2(a) + 9d] 
[2(a) + 9d] = 125/5
[2(a) + 9d] = 25 
2a + 9d = 25 --------- equation 2
4) Subtract double of equation 1 from equation 2
2a + 9d = 25
2a + 4d = 30
  ( - )  ( - )     ( - )  
  --------------------------- 
      5d = -  5
d = - 5/5
d = - 1 --------- equation 3
5) Put d = - 1 from equation 3 in equation 1, and we get,
a + 2d = 15
a + 2(- 1) = 15
a - 2 = 15
a = 15 + 2
a = 17 --------- equation 4 
6) We know that,
an = a + (n – 1) d
a10 = 17 + (10 – 1) (- 1) 
a10 = 17 + 9 (- 1)
a10 = 17 - 9
a10 = 8
7) So, a10 = 8, d = - 1. 
 
(v) given d = 5, S9 = 75, find a and a9.

1) Here, d = 5, S9 = 75. Find a and a9.
2) We know that,
Sn = (n/2)[2a + (n - 1) d]
S9 = (9/2)[2a + (9 - 1) (5)]
75 = (9/2)[2a + 8 (5)]
75 = (9/2)[2a + 40]
75 = 9[a + 20]
[a + 20] = 75/9
[a + 20] = 25/3
a = (25/3) - 20
a = (25 - 60)/3
a = - 35/3
3) We know that the 
an = a + (n – 1) d
a9 = - 35/3 + (9 – 1) (5)
a9 = - 35/3 + (8)(5)
a9 = - 35/3 + 40
a9 = (- 35 + 120)/3
a9 = 85/3
4) So, a9 = 85/3, a = - 35/3.

(vi) given a = 2, d = 8, Sn = 90, find n and an.

1) Here, a = 2, d = 8, Sn = 90. Find n and an.
2) We know that,
Sn = (n/2)[2a + (n - 1) d]
Sn = (n/2)[2(2) + (n - 1)(8)]
90 = (n/2)[4 + 8(n - 1)]
90 = n[2 + 4(n - 1)]
90 = n(2 + 4n - 4)
90 = n(4n - 2)
90 = 4n2 - 2n
45 = 2n2 - n
2n2 - n - 45 = 0
2n2 - 10n + 9n - 45 = 0
2n(n - 5) + 9(n - 5) = 0
(n - 5)(2n + 9) = 0
(n - 5) = 0 or (2n + 9) = 0
n = 5 or 2n = - 9
n = 5 or n = - 9/2 
3) As n is always a positive integer, ignore n = - 9/2. So we have n = 5.
4) We know that,
an = a + (n – 1) d
a5 = 2 + (5 – 1) (8)
a5 = 2 + (8)(4)
a5 = 2 + 32
a5 = 34
5) So, a5 = 34, n = 5. 

(vii) given a = 8, an = 62, Sn = 210, find n and d.

1) Here, a = 8, an = 62, Sn = 210. Find n and d.
2) We know that,
Sn = (n/2)[a + l]
210 = (n/2)[8 + 62]
210 = (n/2)[70]
210 = 35n
35n = 210
n = 210/35
n = 6
 3) Now we will find d using the formula:
an = a + (n – 1) d
62 = 8 + (6 – 1) (d)
62 = 8 + 5d
5d = 62 - 8
5d = 54
d = 54/5
d = 10.8
5) So, d = 10.8, n = 6. 

(viii) given an = 4, d = 2, Sn = – 14, find n and a.

1) Here, d = 2, an = 4, Sn = - 14. Find n and a.
2) We know that,
Sn = (n/2)[a + l]
- 14 = (n/2)[a + 4] 
(n/2)[a + 4] = - 14
n[a + 4] = - 14(2)
n[a + 4] = - 28 --------- equation 1
 3) We know that,
an = a + (n – 1) d
4 = a + (n – 1) (2)
4 = a + 2n – 2
a + 2n = 4 + 2
a + 2n = 6
a = 6 - 2n --------- equation 2
4) Put a = 6 - 2n from equation 2 in equation 1, we get, 
n(a + 4) = - 28
n((6 - 2n) + 4) = - 28
n(6 - 2n + 4) = - 28
n(10 - 2n) = - 28 
2n(5 - n) = - 28
n(5 - n) = - 14
5n - n2 = - 14
n2 - 5n - 14 = 0
n2 - 7n + 2n - 14 = 0
n(n - 7) + 2(n - 7) = 0
(n - 7)(n + 2) = 0
(n - 7) = 0 or (n + 2) = 0
n = 7 or n = - 2
5) As n is always a positive integer, ignore n = - 2. So we have n = 7.
6) Put n = 7 in equation 2, and we get,
a = 6 - 2n
a = 6 - 2(7)
a = 6 - 14
a = - 8
7) So, a = - 8, n = 7.

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

1) Here, a = 3, n = 8, sum of first n terms is Sn = 192, find d.
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n - 1) d]
192 = (8/2)[2(3) + (8 - 1) d]
192 = 4[6 + 7d]
4[6 + 7d] = 192
[6 + 7d] = 192/4
[6 + 7d] = 48
7d = 48 - 6
7d = 42
d = 42/7
d = 6
3) So, d = 6.
 
(x) given l = 28, S = 144, and there are total 9 terms. Find a.

1) Here, l = 28, Sn = 144, n = 9, Find a.
2) We know that,
Sn = (n/2)[a + l]
144 = (9/2)[a + 28] 
(9/2)[a + 28] = 144
[a + 28] = 144(2/9)
[a + 28] = 16 (2)
a + 28 = 32
a = 32 - 28
a = 4
3) So, a = 4.

Q4. How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of
636?

Solution:

1) Here, a1 = a = 9, a2 = 17, a3 = 25, sum of first n terms is Sn = 636, find n.
2) According to the problem,
d = a2 - a1
d = 17 - 9
d = 8
3) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[2a + (n - 1) d]
636 = (n/2)[2(9) + (n - 1) (8)]
636 = (n/2)[18 + 8(n - 1)]
636 = n[9 + 4(n - 1)]
636 = n[9 + 4n - 4]
636 = n[4n + 5]
636 = 4n2 + 5n
4n2 + 5n - 636 = 0
4n2 + 53n - 48n - 636 = 0
n(4n + 53) - 12(4n + 53) = 0
(4n + 53)(n - 12) = 0
(4n + 53) = 0 or (n - 12) = 0
n = - 53/4 or n = 12
4) As n is always a positive integer, ignore n = - 53/4. So we have n = 12.

Q5. 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.

Solution:

1) Here, a1 = a = 5, an = 45, Sn = 400, find n and d.
2) We know that the sum of the first n terms of an AP is given by:
S = (n/2)[a + l]
400 = (n/2)[(5) + 45]
400 = (n/2)[50]
400 = n[25]
n[25] = 400
n = 400/25
n = 16
3) Now we will find the value of d. 
an = a + (n – 1) d 
45 = 5 + d (16 – 1)
15d = 45 - 5
15d = 40
d = 40/15
d = 8/3 
4) So, d = 8/3, n = 16.

Q6. 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?

Solution:

1) Here, a1 = a = 17, an = 350, d = 9, find n and Sn.
2) Now we will find the value of d. 
an = a + (n – 1) d 
350 = 17 + 9(n – 1)
9(n – 1) = 350 - 17
9(n – 1) = 333
(n – 1) = 333/9
(n – 1) = 37
n = 37 + 1
n = 38 
3) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[a + l]
Sn = (38/2)[(17) + 350]
Sn = 19[17 + 350]
Sn = 19(367)
Sn = 6973
4) So, n = 38, Sn = 6973.

Q7. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.

Solution:

1) Here, a22 = 149, d = 7, find and S22.
2) Now we will find the value of d. 
an = a + (n – 1) d
a22 = a + (22 – 1) (7) 
149 = a + 7(22 – 1)
a = 149 - 7(21)
a = 149 - 147
a = 2 
3) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[a + l]
S22 = (22/2)[(2) + 149]
S22 = (11)[151]
S22 = 1661
4) So, a = 2, Sn = 1661.

Q8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

1) Here, a2 = 14, a3 = 18, find and S51.
2) According to the problem,
d = a3 - a2
d = 18 - 14
d = 4
3) Now we will find the value of d. 
an = a + (n – 1) d
a2 = a + (2 – 1) (4) 
14 = a + 4(1)
a = 14 - 4
a = 10
4) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1)d]
S51 = (51/2)[2(10) + (51 - 1)(4)]
S51 = (51/2)[20 + (50)(4)]
S51 = (51/2)[20 + 200]
S51 = (51/2)[220]
S51 = 51[110]
S51 = 5610
5) So, S51 = 5610.

Q9. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n terms.

Solution:

1) Here, S7 = 49, S17 = 289, find and Sn. Let the first term be 'a' and the common
difference be 'd'. 
2) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1)d]
S7 = (7/2)[2a + (7 - 1)(d)]
49 = (7/2)[2a + (6)(d)]
49 = 7[a + 3d]
a + 3d = 49/7
a + 3d = 7 --------- equation 1
3) Now we will use S17 = 289 to get another equation
Sn = (n/2)[2a + (n - 1)d]
S17 = (17/2)[2a + (17 - 1)d]
S17 = (17/2)[2a + (16)d]
S17 = 17[a + 8d]
289 = 17[a + 8d]
17[a + 8d] = 289
a + 8d = 289/17
a + 8d = 17 --------- equation 2
4) Subtract equation 1 from equation 2
a + 8d = 17
a + 3d =   7
     ( - )  ( - )     ( - )
--------------------------
    5d = 10
d = 10/5
d = 2 --------- equation 3
5) Put d = 2 from equation 3 in equation 1, and we get,
a + 3d = 7
a + 3(2) = 7
a + 6 = 7
a = 7 - 6
a = 1
2) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1)d] 
Sn = (n/2)[2(1) + (n - 1)(2)]
Sn = n[1 + (n - 1)]
Sn = n[n]
Sn = n2 
5) So, Sn = n2.

Q10. Show that a1, a2, . . ., an, . . . form an AP where an is defined as below :
(i) an = 3 + 4n     (ii) an = 9 – 5n. 
Also, find the sum of the first 15 terms in each case.

Solution:

(i) an = 3 + 4n.

1) Put n = 1, 2, 3 in equation an = 3 + 4n to get a1, a2, a3  and so on.
a) Put n = 1,
an = 3 + 4n
a1 = 3 + 4(1)
a1 = 3 + 4
a1 = 7
b) Put n = 2,
an = 3 + 4n
a2 = 3 + 4(2)
a2 = 3 + 8
a2 = 11
c) Put n = 3,
an = 3 + 4n
a3 = 3 + 4(3)
a3 = 3 + 12
a3 = 15 
2) Here, a1 = 7, a2 = 11, a3 = 15.
a) First difference:
d = a2 - a1
d = 11 - 7
d = 4 --------- equation 1
b) Second difference:
d = a3 - a2
d = 15 - 11
d = 4 --------- equation 2
3) Here the common difference d = a2 - a1 = a3 - a2 = 4, so the equation an = 3 + 4n
form an AP and their terms are 7, 11, 15 . . .
4) Now we will find the sum of the first 15 terms.
5) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n – 1) d]
S15 = (15/2)[2(7) + (15 – 1) (4)]
S15 = (15/2)[2(7) + 4(14)]
S15 = (15)[7 + 2(14)]
S15 = (15)[7 + 28] 
S15 = (15)(35)
S15 = 525
 6) So, here Sn = 525.

(ii) an = 9 – 5n.

1) Put n = 1, 2, 3 in equation an = 9 - 5n to get a1, a2, a3  and so on.
a) Put n = 1,
an = 9 - 5n
a1 = 9 - 5(1)
a1 = 9 - 5
a1 = 4
b) Put n = 2,
an = 9 - 5n
a2 = 9 - 5(2)
a2 = 9 - 10
a2 = - 1
c) Put n = 3,
an = 9 - 5n
a3 = 9 - 5(3)
a3 = 9 - 15
a3 = - 6
2) Here, a1 = 4, a2 = - 1, a3 = - 6.
a) First difference:
d = a2 - a1
d = - 1 - 4
d = - 5 --------- equation 1
b) Second difference:
d = a3 - a2
d = -  6 - (-1)
d = -  6 + 1 
d = - 5 --------- equation 2
3) Here the common difference d = a2 - a1 = a3 - a2 = - 5, 
so the equation an = 9 - 5n form an AP and their terms are 4, - 1, - 6 . . .
4) Now we will find the sum of the first 15 terms.
5) We know that the sum of the first n terms of an AP is given by:
Sn = (n/2)[2a + (n – 1) d]
S15 = (15/2)[2(4) + (15 – 1) (- 5)]
S15 = (15/2)[2(4) - 5(14)]
S15 = (15)[4 - 5(7)]
S15 = (15)[4 - 35] 
S15 = (15)[- 31] 
S15 = - 465
 6) So, here Sn = - 465.
 
Q11. 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 the first two terms? What is the second term? Similarly, find the 3rd, the 10th, and the nth terms.

Solution:

1) Here Sn = 4n – n2 so,
a) So, first term  S1 will be,
Sn = 4n – n2
S1 = 4(1) – (1)2
S1 = 4 – 1
S1 = 3 = a1 = a
b) To get sum of first two terms, put n = 2, in Sn = 4n – n2
Sn = 4n – n2
S2 = 4(2) – (2)2
S2 = 8 – 4
S2 = 4
2) We know that the second term a2 = S2 - S1.
a2 = S2 - S1
a2 = 4 - 3
a2 = 1
3) The common difference will be:
d = a2 - a1
d = 1 - 3
d = - 2
4) Now we will find 3rd term:
an = a + (n – 1) d
a3 = 3 + (3 – 1) (- 2)
a3 = 3 + (2) (- 2)
a3 = 3 - 4
a3 = - 1
5) Now we will find the 10th term:
an = a + (n – 1) d
a10 = 3 + (10 – 1) (- 2)
a10 = 3 + (9) (- 2)
a10 = 3 - 18
a10 = - 15
6) Now we will find the nth term:
an = a + (n – 1) d
an = 3 + (n – 1) (- 2)
an = 3 - 2n + 2
an = 5 - 2n
7) So here we have:
a) First term  S1 = a = 3.
b) Sum of first two terms S2 = 4. 
c) The second term a2 = 1
d) The third term a3 = - 1
e) The 10th term a10 = - 15
f) The nth term an = 5 - 2n
 
Q12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

1) Here a1 will be 6, and a2 will be 12, so here d = 6.
So, now we will find sum of first 40 integers,
Sn = (n/2)[2a + (n – 1) d
S40 = (40/2)[2(6) + (40 – 1) (6)]
S40 = 20[12 + 6(39)]
S40 = 20[12 + 234]
S40 = 20[246]
S40 = 4920
2) So the sum of the first 40 terms will be 4920.

Q13. Find the sum of the first 15 multiples of 8.

Solution:

1) Here a1 will be 8, and a2 will be 16, so here d = 8.
So, now we will find sum of first 15 multiples of 8,
Sn = (n/2)[2a + (n – 1) d
S15 = (15/2)[2(8) + (15 – 1) (8)]
S15 = 15[8 + 4(14)]
S15 = 15[8 + 56]
S15 = 15[64]
S15 = 960
2) So the sum of the first 15 multiples of 8 will be 960.

Q14. Find the sum of the odd numbers between 0 and 50.

Solution:

1) Here, our odd numbers are 1, 3, 5, . . 47, 49.
2) We have a1 = 1, a2 = 3, d = 2 and an = 49,
an = a + (n – 1) d
49 = 1 + 2(n – 1)
2(n – 1) = 49 - 1
2(n – 1) = 48
(n – 1) = 48/2
(n – 1) = 24 
n = 25
 3) According to the problem,
So, now we will find sum of odd numbers from 0 to 50.
Sn = (n/2)[a + l]
S25 = (25/2)[1 + 49]
S25 = (25/2)[50]
S25 = 25[25]
S25 = 625
4) The sum of all odd numbers between 0 and 50 is 625.

Q15. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days?

Solution:

1) Here a1 will be 200, and a2 will be 250, so here d = 50.
So, now we will find total penalty for the delayed work by 30 days, so we will have to find Sn.
Sn = (n/2)[2a + (n – 1) d
S30 = (30/2)[2(200) + (30 – 1) (50)]
S30 = 15[400 + 50(29)]
S30 = 15[400 + 1450]
S30 = 15[1850]
S30 = 27750
2) So the total penalty for the delayed work by 30 days is 27750.

Q16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Solution:

1) Let the cost of the first prize be Rs x.
2) According to the problem, the cost of the second prize is (x - 20).
3) The cost of the third prize is (x - 40). So the common difference is d = - 20.
4) As the sum of Rs 700 is to be used give 7 cash prizes, S7 = 700.
So, now we will find total penalty for the delayed work by 30 days, so we will have to find Sn.
Sn = (n/2)[2a + (n – 1) d
S7 = (7/2)[2(x) + (7 – 1) (- 20)]
S7 = (7)[(x) + (6) (- 10)]
700 = (7)[(x) - 60]
100 = x - 60
x = 100 + 60
x = 160 
5) So, the value of each of the prizes was (i) Rs 160, (ii) Rs 140, (iii) Rs 120, 
(iv) Rs 100, (v) Rs 80, (vi) Rs 60, and (vii) Rs 40.

Q17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

1) Here a1 will be 1, and a2 will be 2, an will be 12, so here d = 1.
So, now we will find total penalty for the delayed work by 30 days, so we will have to find Sn.
Sn = (n/2)[2a + (n – 1) d
S12 = (12/2)[2(1) + (12 – 1) (1)]
S12 = 6[2 + 1(11)]
S12 = 6[2 + 11]
S12 = 6[13]
S12 = 78
2) So one section of a school will plant 78 trees. So 3 sections will plant 
78 x 3 = 234 plants.

Q18. A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take p = 22/7)

Solution:

1) We know that the perimeter of the first semicircle is 𝞹r.
2) According to the problem, the perimeters of the semicircle are:
a) r = 0.5 so  a1 = 0.5 𝞹 = 𝞹/2
b) r = 1 so  a2 = 𝞹 = 𝞹
c) r = 1.5 so  a3 = 1.5 𝞹 = 3𝞹/2
3) Here a1, a2, and a3, are the semicircle's perimeter and are in AP.
4) Here d = a2 - a1 = 𝞹 - (𝞹/2) = (𝞹/2)
5) We will have to find S13, so we have,
Sn = (n/2)[2a + (n – 1) d
S13 = (13/2)[2(𝞹/2) + (13 – 1) (𝞹/2)]
S13 = (13/2)(𝞹/2)[2 + 12]
S13 = (13/2)(𝞹/2)[14]
S13 = (13/2)(𝞹)[7]
S13 = (13/2)(22/7)[7]
S13 = (13/2)(22)
S13 = (13)(11)
S13 = 143
6) So, the length of such a spiral of thirteen consecutive semicircles is 143 cm.

Q19. 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?

Solution:

1) Here a number of logs in the rows are in an AP.
2) According to the problem, a1 = 20, a2 = 19, a3 = 18. . . and Sn = 200, so d = - 1,
3) We will have to find n using the above information.
Sn = (n/2)[2a + (n – 1) d
Sn = (n/2)[2(20) + (n – 1) (- 1)]
200 = (n/2)[40 - (n - 1)]
200 = (n/2)[40 - n + 1)] 
400 = n(41 - n)
400 = 41n -  n2
n2 - 41n + 400 = 0
n2 - 16n - 25n + 400 = 0
n(n - 16) - 25(n - 25) = 0
(n - 16)(n - 25) = 0
(n - 16) = 0 or (n - 25) = 0
n = 16 or n = 25
4) Now we will find a16 and a25 using the formula an = a + (n – 1) d. 
a) First we will find a16 
an = a + (n – 1) d 
a16 = 20 + (- 1) (16 – 1)
a16 = 20 - (15)
a16 = 5
b) First we will find a25 
an = a + (n – 1) d 
a25 = 20 + (- 1) (25 – 1)
a25 = 20 - (24)
a25 = - 4
5) As the number of logs can't be negative, a25 = - 4 is impossible. So 200 logs
can be placed in 16 rows and there are 5 logs in the 16th log.

Q20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first potato and the second potato, the total distance (in meters) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Solution:

1) The potatoes are placed at 5, 8, 11, 14, 17 . . . distances from the bucket.
2) These distances form an AP.
3) According to the problem, a competitor travels distances as 10, 16, 22, 28, . . .
4) So a1 = 10, a2 = 16, a3 = 22, a4 = 28. . . , so d = 6,
5) Now we will have to find S10 using the above information.
Sn = (n/2)[2a + (n – 1) d
S10 = (10/2)[2(10) + (10 – 1) (6)]
S10 = (10/2)(2)[10 + (9) (3)]
S10 = 10[10 + 27] 
S10 = 10[37]
S10 = 370
6) A competitor will have to run a total distance of 370 m.

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