Telangana SCERT Solution Class X (10) Maths Chapter 6 Progressions Exercise 6.3
(Q1) Find the sum of the following A.P.s
(i) 2, 7, 12, —– to 10 terms
=> Solution:
Given A.P is 2, 7, 12, ——-
A = 2, d = a2 – a1 = 7 – 2 = 5
And n = 10
We have, Sn = n/2 [2a + (n – 1) d]
S10 = 10/2 [2 X 2 + (10 – 1) 5]
= 5 [4 + (9) 5]
= 5 [4 + 45]
= 5 [49]
S10 = 245
(ii) – 37, – 33, – 29, —— +0 11 terms
=> Solution:
Given A.P is,
-37, -33, -29, —–
a = – 37, d = a2 – a1 = -33 – (-37) = -33 + 37 = 4
and n = 12
We have,
Sn = n/2 [2a + (n – 1) d]
S12 = 12/2 [2 X (-37) + (12 – 1) (4)]
= 6 [-74 + (11) (4)]
= 6 [-74 + 44]
= 6 [-30]
S12 = – 180
(iii) 0.6, 1.7, 2.8, —– to 100 terms
=> Solution:
Given A.P. is,
0.6, 1.7, 2.8, —–
A = 0.6, a = a2 – a1 = 1.7 – 0.6 = 1.1
and n = 100
We have, Sn = n/2 [2a + (n – 1) d]
S100 = 100/2 [2 X (0.6) + (100 – 1) (1.1)]
= 50 [1.2 + (99) 1.1]
= 50 [1.2 + 108.9]
= 50 [110.1]
S100 = 5, 505
(iv) 1/15, 1/12, 1/10, —-, + 0 11 terms
=> Solution:
Given A.P is,
1/15, 1/12, 1/10, —–
a = 1/15, d = a2 – a1 = 1/12 – 1/15
= 5-4/60
= 1/60
and n = 11
We have, Sn = n/2 [2a + (n – 1) d]
S11 = 11/2 [2 X 1/15 + (11 – 1) 1/60]
= 11/2 [2/15 + (10) 1/60]
= 11/2 [2/15 + 1/6]
= 11/2 [4+5/30]
= 11/2 [9/30]
= 99/60
Sn = 33/20
(Q2) Find the sums given below:
(i) 7 + 10 1/2 + 14 + —– + 84
=> Solution:
d = a2 – a1
= 10 1/2 – 7
= 21/2 – 7
d = 21-14/2 = 7/2
a3 – a2 = 14 – 10 1/2 = 14 – 21/2 = 28-21/2 = 7/2
It forms an A.P
Let an = 84
a + (n – 1) d = 84
7 + (n – 1) 7/2 = 84
(n – 1) 7/2 = 84 – 7
(n – 1) 7/2 = 77
n – 1 = 77 X 2/7
= 11 X 2
n – 1 = 22
n = 22 + 1
n = 23
We have,
Sn = n/2 [2a + (n – 1) d]
S23 = 23/2 [2 X 7 + (23 – 1) 7/2]
= 23/2 [14 + (22) 7/2]
= 23/2 [14 + 11 X 7]
= 23/2 [14 + 77]
= 23/2 [91]
S23 = 2,093/2
S23 = 1046 1/2
(ii) 34 + 32 + 30 + —– + 10
=> Solution:
d = a2 – a1 = 32 – 34 = – 2
a3 – a2 = 30 – 32 = -2
d = -2
It forms an A.P
Let an = 10
We have,
a + (n – 1) d = 10
34 + (n – 1) (-2) = 10
34 + (n – 1) (-2) = 10
(n – 1) (-2) = 10 – 34
(n – 1) (-2) = – 24
n – 1 = – 24/2
n – 1 = +12
n = + 12 + 1
n = + 13
We have,
Sn = n/2 [2a + (n – 1) d]
S13 = 13/2 [2 X 34 + (13 – 1) (-2)]
= 13/2 [68 + (12) (-2)]
S13 = 13/2 [68 – 24]
= 13/2 [44]
= 13/2 X 22
= 572/2
S13 = 286
(iii) – 5 + (-8) + (-11) + —- + (-230)
=> Solution:
d = a2 – a1 = -8 – (-5) = -8 + 5 = -3
a3 – a2 = – 11 – (-8) = – 11 + 8 = – 3
It forms an A.P
Let an = – 230
a + (n – 1) d = – 230
– 5 + (n – 1) (-3) = – 230
(n – 1) (-3) = – 230 + 5
(n – 1) (-3) = -225
n – 1 = -225/-3
n – 1 = 75
n = 75 + 1
n = 76
We have,
Sn – 1/2 [2a + (n – 1) d]
S76 = 76/2 [2 X (-5) + (76 – 1) (-3)]
= 38 [-10 + (75) (-3)]
= 38 [-10 – 225]
= 38 [-235]
S76 = – 8930
(Q3) In an A.P
(i) Given a = 5, d = 3, an = 50, find n and Sn
=> Solution: given, a = 5, d = 3, an = 50
We have,
an = a + (n – 1) d
50 = 5 + (n – 1) 3
50 – 5 = (n – 1) 3
45 = (n – 1) 3
45/3 = (n – 1)
15 = (n – 1)
15 + 1 = n
16 = n
∴ n = 16
We have, Sn = n/2 [2a + (n – 1) d]
= 16/2 (2 X 5 + (16-1) 3]
= 8 [10 + (15)3]
= 8 [10 + 45]
Sn = 8 [5.5]
Sn = 440
(ii) Given a = 7, a13 = 35 find d and 513
=> Solution: Given a = 7, a13 = 35
∴ a13 = 35
a + 12d = 35
7 + 12d = 35 [∵ a = 7]
12d = 35 – 7
12d = 28
d = 28/12
d = 7/3
S13 = n/2 (2a + (n -1) d]
= 13/2 [2 X 7 + (13 – 1) (7/3)]
= 13/2 [14 + (12) (7/3)]
= 13/2 [14 + 4(7)]
S13 = 13/2 [14 + 28]
= 13/2 [42]
= 13 [21]
S13 = 273
∴ d = 7/3 and S13 = 273
(iii) Given a12 = 37, d = 3, find a and s12
=> Solution:
Given a12 = 37, d = 3
a12 = 37
a + 11d = 37
a + 11 X 3 = 37 (∵ d = 3)
a + 33 = 37
a = 37 – 33
a = 4
We have,
Sn = n/2 [2a + (n – 1) d]
S12 = 12/2 [2 X 4 + (12 – 1) 3]
= 6 [8 + 11 X 3]
= 6 [8 + 33]
S12 = 6 [41]
S12 = 246
∴ a = 4 and S12 = 246
(iv) Given a3 = 15, S10 = 125, find d & a10
=> Solution:
Given a3 = 15, S10 = 125
a3 = 15
a + 2d = 15 ——- (1)
S10 = 125
n/2 [2a + (n – 1) d] = 125
10/2 [2a + (10 – 1) d] = 125
5 [2a + (9) d] = 125
2a + 9d = 25 —— (2)
Multiply equation (1) by ‘2’
2 (a + 2d) = 2 X 15
2a + 4d = 30 ——- (3)
Since n is number of terms it must be positive integer.
∴ n = 5
an = a + (n – 1) d
a5 = 2 + (5 – 1) 8
= 2 + (4) 8
= 2 + 32
a5 = 34
∴ n = 5 and a5 = 34
(vi) Given an = 4, d = 2, sn = -14 find n and a
=> Solution:
Given an = 4, d = 2, Sn = -14
an = 4
a + (n – 1) d = 4
a + (n – 1) 2 = 4 (∵ d = 2)
a + 2n – 2 = 4
a + 2n = 4 + 2
a + 2n = 6 —– (i)
Sn = – 14
n/2 [a + an] = -14
n/2 [a + 4] = -14
(vii) Given l = 28, s = 144, and there are total 9 terms, find a
=> Solution:
Given l = 28, s = 144
There are total 9 terms i.e. n =?
Sn = 144
n/2 [a + l) = 144
9/2 [a + 28] = 144
a + 28 = 144 X 2/9
a + 28 = 32
a = 32 – 28
a = 4
(Q4) The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
=> Solution: In the given A.P.
a = 17 and l = 350
d = 9 (∵ Given)
∴ l = 350
a + (n – 1) = 350
17 + (n – 1) 9 = 350
(n – 1) 9 = 350 – 17
9 (n – 1) = 333
n – 1 = 333/9
n – 1 = 37
n = 37 + 1
n = 38
We have, Sn = n/2 [a + l]
S38 = 38/2 [17 + 350]
= 19 [367]
S38 = 6973
∴ There are 38 terms and their sum = 6973.
S51 = 51/2 [2 X (10) + (51 – 1) 4]
= 51/2 [20 + (50) 4]
= 51/2 [20 + 200]
= 51 [220]
S51 = 5610
(Q6) If the sum of first 7 terms of an A.P is 49 and that of 17 terms is 189, find the sum of first n terms
=> Solution:
Sum of the first 7 terms is 49
We have, Sn = n/2 [2a + (n – 1) d]
S7 = 7/2 [2a + (7 – 1) d] = 49
7/2 [2a + 6d] = 49
7/2 X 2 [a + 3d] = 49
7 [a + 3d] = 49
a + 3d = 49/7
a + 6 = 7
a = 7 – 6
a = 1
We have,
Sn = n/2 [2a + (n – 1) d]
Sn = n/2 [2 X 1 + (n – 1) 2]
= n/2 [a + 2n – 2]
= n/2 [2n]
Sn = n2
∴ The sum of the first ‘n’ terms,
Sn = n2
(Q7) Show that a1, a2, —- an, form an A.P. 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 (Given)
If n = 1 à a1 = 3 + 4 (1) = 3 + 4 = 7
If n = 2 à a2 = 3 + 4 (2) = 3 + 8 = 11
If a = 3 => a3 = 3 + 4(3) = 3 + 12 = 15
a1, a2, a3, —– = 7, 11, 15, —
d = a2 – a1 = 11 – 7 = 4
a3 – a2 = 15 – 11 = 4
∴ If forms an A.P.
∴ a = 7 and d = 4
And the sum of the first 15 terms = 515
We have,
5n = n/2 [2a + (n – 1) d]
Here, n = 15, a = 7 and d = 4
S15 = 15/2 [2 (7) + (15 – 1) 4]
= 15/2 [14 + (14) 4]
= 15/2 [14 + 56]
= 15/2 [70]
= 15 (35)
515 = 525
(ii) a = 9 – 5n (∵ Given)
If n = 1 then a1 = 9 – 5(1) = 4
If n = 2 then a2 = 9 – 5(2) = 9 – 10 = – 1
If n = 3 then a3 = 9 – 5(3) = 9 – 15 = – 6
a1, a2, a3 = 4, -1, -6, —
d = a2 – a1 = -1 – 4 = – 5
a3 – a2 = -6 – (-1) = -6 + 1 = – 5
∴ It forms an A.P.
a = 4 and d = -5
And the sum of the first 15 terms = s
We have, Sn = n/2 [2a + (n – 1) d]
Here, n = 15, a = 4, and d = -5
S15 = 15/2, [2 X 4 + (15 – 1) (-5)]
= 15/2 [8 + (14) (-5)]
= 15/2 [8 + (14) (-5)]
= 15/2 [8 – 70]
= 15/2 [-62]
= 15 (-31)
S15 = – 465
(Q8) If the sum of the first n terms of an A.P. is 4n – n2, what is the first term (note that the first term 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
=> Solution:
In an A.P
Sn = 4n – n2 ——- (1)
If n = 1 => S1 = 4(1) – (1)2
= 4 – 1 = 3
If n = 2 => S2 = 4(2) – 22
= 8 – 4 = 4
If n = 3 => S3 = 4(3) – 32
= 12 – 9 = 3
The first term a1 = s1 = 3
Sum of the two terms S2 = 4
Second term a2 = S2 – S1
= 4 – 3
Third term a3 = S3 – S2
= 3 – 4
a3 = -1
Tenth term, a10 = S10 – S9
S10 = 4(10) – 102 = 40 – 100 = – 60 (from (1))
S9 = 4(9) – 92 = 36 – 81 = -45 (∵ from (1))
∴ a10 = S10 – S9
= – 60 – (-45)
= – 60 + 45
a10 = – 15
The nth term an = Sn – Sn – 1
∴ Sn – 1 = 4 (n – 1) – (n – 1)2
= 4n – 4 – [n2 – 2n + 1]
= 4n – 4 – n2 + 2n – 1
= 6n – 5 – n2
Sn – 1 = n2 – 6n + 5
∴ an = 5n – 5n – 1
= (4n – n2) – (6n – n2 – 5)
= 4n – n2 – 6n + n2 + 5
An = 5 – 2n
∴ The first term = 3
Sum of the two terms = 4
Second term = 1
Third term = – 1
10th term = – 15 and
The nth term = 5 – 2n
(Q9) Find the sum of the first positive integers divisible by 6
=> Solution: The first 40 positive integers divisible by 6 are,
6, 12, 18, —– 240
d = a2 – a1 = 12 – 6 = 6
a3 – a2 = 15 – 6 = 6
It forms an A.P
Here, a = 6, an = 240
and d = 6
we have, Sn = n/2 [a + an]
We have n = 40
S40 = 40/2 [6 + 240]
= 20 [246]
S40 = 4920
∴ The sum of the first 40 positive integers divisible by 6 = 4920
(Q10) 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:
Let the amount of first each prize be RS ‘x’
Then the 2nd, 3rd, 4th, 5th, 6th and the 7th prize will be RS (x – 20), RS (x – 40), RS (x – 60), RS (x – 80), RS (x – 100) and RS (x – 120)
List of prizes are,
x, x – 20, x – 40, x – 60, x – 80, x – 100, x – 120
d = a2 – a1 = x – 20 – x = – 20
a3 – a2 = x – 40 – (x – 20)
= x – 40 – x + 20
= – 40 + 20
a3 – a2 = – 20
∴ it forms an A.P
Here, a = x and an = x – 120
We have, Sn = n/2 [a + an]
Since the sum of the 7 prizes is RS 700
I.e. S7 = 100
S7 = 7/2 (x + x – 120) = 700
= 7/2 ( – 120 + 2x) = 700
7/2 (2x – 120) = 700
7 (2x – 120) = 700 X 2
2x – 120 = 700X2/7
= 100 X 2
2x – 120 = 200
2 (x – 60) = 200
x – 60 = 200/2
x – 60 = 100
x = 100 + 60
x = 160
∴ x – 20 = 160 – 20 = 140
x – 40 = 160 – 40 = 120
x – 60 = 160 – 60 = 100
x – 80 = 160 – 80 = 80
x – 100 = 160 – 100 = 60
x – 120 = 160 – 120 = 40
∴ The amount of 7 each prizes are RS 160, RS 140, RS 120, RS 100, RS 80, RS 60 and RS 40
(Q11) Students of a school decided to plant 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 soon till class (XII). There are three sections of each class. How many trees will be planted by the students?
=> Solution:
There are totally 12 class in the school like, (I), (II), (III), (XII)
The list of plants planted by each class will be 1, 2, 3, 4, 5, —— 12
But there are ‘3’ sections in each class then the list of plants planted will be,
3, 6, 9, 12, 15, —– 36
d = a2 – a1 = 6 – 3 = 3
a3 – a2 = 9 – 6 = 3
∴ It forms an A.P.
=> Solution:
Given that, A spiral is made up of successive semicircles, with centres alternately at ‘A’ and ‘B’ with first semi circle centered at ‘A’
Therefore, for us to get 13 semi circles there will be 7 semi circles centered at ‘A’ & 6 at ‘B’
It can be seen that radius with respect to A and B are in A.P with constant difference.
d = 1 for A and B
Given: 0.5, 1.0, 1.5, 2.0, —-
a = 0.5 for a
Given: a7 = a + 6d
a7 = 0.5 + 6d
a7 = 0.5 + 6(1)
= 0.5 + 6
a7 = 6.5
∴ Sum of all the rad of the semi circles centered at A
We have,
Sn = n/2 (a + an)
Here, n = 7, a = 0.5, a7 = 6.5
S7 = 7/2 (0.5 + 6.5)
= 7/2 (7)
S7 = 49/2
S7 = 24.5
∴ a = 1 for B
a6 = a + 5d
= 1 + 5(1)
= 1 + 5
a6 = 6
Sum of all the radii of the semi circles centered at B
We have,
Sn = n/2 (a + an)
Here, n = 6, a = 1, an = 6
S6 = 6/2 (1 + 6)
= 6/2 (7)
= 3 (7)
S6 = 21
∴ Length of 1 spiral = circumference of 1 semi circle
∴ Length of 7 spiral entered at A
= circumference of 7 semi circles centered at A
= Pie X sum of all radii
= 22/7 X 49/2 (∵ π = 22/7)
∴ = 1078/14
= 77 cm
∴ Length of 7 spiral centered at A = 77 cm ——- (i)
∴ Length of 7 spirals centered at B
= Circumference of 6 semi circles centered at B
= Pie X sum of all radii
= 22/7 X (21) (∵ π = 22/7)
= 22 X 3
= 66 cm
∴ Length of 7 spirals centered at B = 66 cm —— (2)
Therefore,
Length of 13 spirals = equn (1) + equn (2)
= 77 + 66
= 143 cms.
∴ Length of 13 spirals = 143 cms.
(Q13) 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:
Given: 20 logs in the bottom row, 19 in the next row, 18 in the next row and soon.
∴ List of the A.P is,
20, 19, 18,
∴ a = 20, d = a2 – a1 = 19 – 20 = – 1
and Sn = 200
We have, 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)
200 = n/2 (41 – n)
2 X 200 = n (41 – n)
Same way until all the balls are in the bucket. What is the total distance the competitor has to run?
[Hint: To pick up the first ball and the second ball, the total distance (in meters) run by a competitor is 2 X 5 + 2 X (5 + 3)]
=> Solution:
Given that, competitor’s pick up the first ball,
The distance run = 2 X 5
= 10
Competitor pick up the second ball
The distance run = 2 X (5 + 3)
= 2 X (8)
= 16
Ans so on.
List of the A.P is
10, 16, 22, 28, 34 —– 64
We have, Sn = n/2 [2a + (n – 1) d]
Here, n = 10, a = 10, d = a2 – a1
= 16 – 10
d = 6
S10 = 10/2 [2 X 10 + (10 – 1) 6]
= 5 [20 + (9) 6]
= 5 [20 + 54]
= 5 [74]
S10 = 370
∴ The total distance the competitor has to run is 370m.
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