ML Aggarwal Arithmetic and Geometric Progression Exe-9.3 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-9.3 Questions for Arithmetic and Geometric Progression (AP GP) as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
ML Aggarwal Arithmetic and Geometric Progression (AP GP) Exe-9.3 Class 10 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-9 | Arithmetic and Geometric Progression |
| Writer / Book | Understanding |
| Topics | Solutions of Exe-9.3 |
| Academic Session | 2024-2025 |
Arithmetic and Geometric Progression (AP GP) Exe-9.3
ML Aggarwal Class 10 ICSE Maths Solutions
Page-180
Question-1 Find the sum of the following A.P.s :
(i) 2, 7, 12, … to 10 terms
(ii) 1/15, 1/12, 1/10, … to 11 terms
Answer- 1
(i) 2, 7, 12, … to 10 terms
Here a = 2, d = 7 – 2 = 5 and n = 10
S10 = n/2(2a + (n – 1)d)
= 10/2 ((2 × 2) + (10 – 1)5)
= 5(4 + 45), = 5(49), = 245
(ii) First term a = 1/15
Then, d = 1/12 – 1/15
= (5 – 4)/60 , = 1/60
common difference d = 1/60, n = 11
S11 = 11/2(2a + (n – 1)d)
= 11/2 ((2 × (1/15)) + (11 – 1)(1/60))
= 11/2 ((2/15) + (10/60))
= 11/2 (2/15 + 1/6)
= 11/2 (4 + 5)/30
= 11/2 (9/30)
= 11/2(3/10)
= 33/20
Question-2 Find the sums given below :
(i) 34 + 32 + 30 + … + 10
(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)
Answer -2
(i) 34 + 32 + 30 + … + 10
Here, a = 34, d = 32 – 34 = -2, l = 10
Tn = a + (n – 1)d
10 = 34 + (n – 1)(-2)
-24 = -2 (n – 1)
-24 = – 2n + 2
2n = 24 + 2
2n = 26, n = 26/2, n = 13
Sn = n/2(a + 1)
= 13/2 (34 + 10)
= 13/2 (44)
= 13 (22)
= 286
(ii) First term a = -5,
Difference d = -8 – (-5) = -8 + 5 = -3
So, common difference d = – 3
Last term Tn = -230
We know that, Tn = a + (n – 1)d
-230 = -5 + (n – 1)(-3)
-230 = – 5 – 3n + 3
-230 = – 2 – 3n
3n = 230 -2
3n = 228
n = 228/3
n = 76
Therefore, Sn = n/2 (a + l)
= 76/2 (-5 + (-230))
= 38 (-5 – 230)
= 38 (235)
= – 8930
Question-3 In an A.P. (with usual notations) :
(i) given a = 5, d = 3, an = 50, find n and Sn
(ii) given a = 7, a13 = 35, find d and S13
(iii) given d = 5, S9 = 75, find a and a9
(iv) given a = 8, an = 62, Sn = 210, find n and d
(v) given a = 3, n = 8, S = 192, find d.
Answer-3
(i) a = 5, d = 3, an = 50
an = a + (n – 1 )d
50 = 5 + (n – 1)3
50 = 5 + 3n – 3
50 = 2 + 3n
3n = 50 – 2
3n = 48, n = 48/3, n = 16
So, Sn = (n/2)(2a + (n – 1)d)
= (16/2) ((2 × 5) + (16 – 1) × 3)
= 8(10 + 45), = 8(55), = 440
(ii) First term a = 7
a13 = 35,
We know that, an = a + (n – 1)d
35 = 7 + (13 – 1)d
35 = 7 + 13d – d
35 = 7 + 12d
12d = 35 – 7
12d = 28
d = 28/12 … [divide by 4]
d = 7/3
So, S13 = (n/2)(2a + (n – 1)d)
= (13/2) ((2 × 7) + ((13 – 1) × (7/3))
= (13/2) ((14 + (12 × 7/3))
= (13/2) (14 + 28)
= (13/2) (42) = 273
(iii) Common difference d = 5
S9 = 75
We know that, an = a + (n – 1)d
a9 = a + (9 – 1)5
a9 = a + 45 – 5
a9 = a + 40 … equation (i)
Then, S9 = (n/2) (2a + (n – 1)d)
75 = (9/2) (2a + (9 – 1)5)
75 = (9/2) (2a + (8)5)
(75 × 2)/9 = 2a + 40
150/9 = 2a + 40
2a = 150/9 – 40
2a = 50/3 – 40
2a = (50 – 120)/3
2a = -70/3
a = -70/(3 × 2)
a = – 35/3
, substitute the value of a in equation (i),
a9 =a + 40
= -35/3 + 40
= (-35 + 120)/3
= 85/3
(iv) First term a = 8,
an = 62 and Sn = 210
We know that, an = a + (n – 1)d
62 = 8 + (n – 1)d
(n – 1)d = 62 – 8
(n – 1)d = 54 … [equation (i)]
Then, Sn = (n/2) (2a + (n – 1)d)
210 = (n/2) ((2 × 8) + 54) … [from equation (i) (n – 1)d = 54]
210 = (n/2) (16 + 54)
420 = n(70)
n = 420/70
n = 6
Now, substitute the value of n in equation (i),
(n – 1)d = 54
(6 – 1)d = 54
5d = 54
d = 54/5
Therefore, d = 54/5 and n = 6
(v) First term a = 3
n = 8
S = 192
We know that, Sn = (n/2) (2a + (n – 1)d)
192 = (8/2) ((2 × 3) + (8 – 1)d)
192 = 4 (6 + 7d)
192/4 = 6 + 7d
48 = 6 + 7d
48 – 6 = 7d
42 = 7d
d = 42/7
d = 6
Hence common difference d is 6.
Question-4
(i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Answer -4
(i) First term of an A.P. (a) = 5
Last term (l) = 45
Sum = 400
l = a + (n – 1 )d
45 = 5 + (n – 1)d
⇒ (n – 1)d = 45 – 5 = 40 …(i)
So, Sn = (n/2) (2a + (n – 1)d)
400 = (n/2) ((2 × 5) + 40) … from equation (i) (n – 1)d = 40
800 = n(10 + 40)
800 = 50n
n = 800/50, n = 16
(ii) First term a = 15
Therefore, sum of first n terms of an A.P. is given by,
Sn = (n/2) (2a + (n – 1)d)
S15 = (15/2)(2a + (15 – 1)d)
750 = (15/2) (2a + 14d)
(750 × 2)/15 = 2a + 14d
100 = 2a + 14d
Dividing both the side by 2 we get,
50 = a + 7d
Now, substitute the value a,
50 = 15 + 7d
7d = 50 – 15
7d = 35
d = 35/7
d = 5
So, 20th term a20 = a + 19d
= 15 + 19(5)
= 15 + 95, = 110
Question -5 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?
Answer -5
First term of an A.P. (a) = 17
and last term (l) = 350
d= 9
l = Tn = a + (n – 1 )d
350 = 17 + (n – 1) × 9
350 – 17 = 9n – 9
333 + 9 = 9n
342 = 9n
n = 342/9
n = 38
So, Sn = (n/2) (2a + (n – 1)d)
= (38/2) ((2 × 17) + (38 – 1)d)
= 19(34 + (37 × 9))
= 19(34 + 333)
= 19 × 367
= 6973
Hence n = 38 and Sn = 6973
Arithmetic and Geometric Progression Exe-9.3
ML Aggarwal Class 10 ICSE Maths AP GP Solutions.
Question -6 Solve for x : 1 + 4 + 7 + 10 + … + x = 287.
Answer-6
First term a = 1
Difference d = 4 – 1 = 3
n = x
x = a = (n – 1)d
x – 1 = (n – 1)d
Sn = (n/2) (2a + (n – 1)d)
287 = (n/2) ((2 × 1)+ (n – 1)3)
= n (2 + 3n – 3)
574 = n(2 + 3n – 3)
574 = 2n + 3n2 – 3n
574 = – n + 3n2
3n2 – n – 574 = 0
3n2 – 42n + 41 – 574 = 0
3n(n – 14) + 41(n – 14) = 0
(n – 14) (3n + 41) = 0
If n – 14 = 0
n = 14
or 3n + 41 = 0
3n = -41
n = -41/3
taking positive number so n = 14
Then, = a + (n – 1)d
= 1 + (14 – 1) 3
= 1 + (13)3
= 1 + 39 , = 40
Hence, x = 40
Question -7
(i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.
(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.
Answer-7
(i) A.P. is 25, 22, 19, …
Sum = 116
Here, a = 25, d = 22 – 25 = -3
Let number of terms be n, then
Sn = (n/2) (2a + (n – 1)d)
116 = (n/2) (2a + (n – 1)d)
on cross multiplication,
232 = n ((2 × 25) + (n – 1) (-3))
232 = n (50 – 3n + 3)
232 = n (53 – 3n)
232 = 53n – 3n2
3n2 – 53n + 232 = 0
3n2 – 24n – 29n + 232 = 0
3n (n – 8) – 29 (n – 8) = 0
(n – 8) (3n – 29) = 0
If n – 8 = 0
n = 8
or 3n – 29 = 0
3n = 29
n = 29/3
fraction, not possible
So, n = 8
Then, T = a + (n – 1)d
= 25 +(8 – 1) (-3)
= 25 + 7 (-3)
= 25 – 21, = 4
(ii) First term a = 24
Common difference d = 21 – 24 = – 3
Sum = 78
Sn = (n/2) (2a + (n – 1)d)
78 = (n/2) (2a + (n – 1)d)
on cross multiplication,
156 = n ((2 × 24) + (n – 1) (-3))
156 = n (48 – 3n + 3)
156 = n (51 – 3n)
156 = 51n – 3n2
3n2 – 51n + 156 = 0
3n2 – 12n – 39n + 156 = 0
3n (n – 4) – 39 (n – 4) = 0
(n – 4) (3n – 39) = 0
If n – 4 = 0
n = 4
or 3n – 39 = 0
3n = 39
n = 39/3
n = 13
So, n = 4
Then, T = a + (n – 1)d
= 24 +(4 – 1) (-3)
= 24 + 3 (-3)
= 24 – 9
= 15
n = 13
Then, T = a + (n – 1)d
= 24 +(13 – 1) (-3)
= 24 + 12 (-3)
= 24 – 36
= -12
So, (12 + 9 + 6 + 3 + 0 + (-3)+ (-6) + (-9) + (-12)) = 0
Hence, the sum of 5th term to 13th term = 0
Question-8 Find the sum of first 22 terms, of an A.P. in which d = 7 and a22 is 149.
Answer-8
Sum of first 22 terms of an A.P. whose d = 7
a22 = 149 and n = 22
a22 = (n – 1)d
149 = a + (22 – 1)7
149 = a + (22)7
149 = a + 147
a = 149 – 147
a = 2
So, S22 = (n/2) (2a + (n – 1)d)
= (22/2) ((2 × 2) + (22 – 1)7)
= 11(4 + (21)7)
= 11 (4 + 147)
= 11 (151)
= 1661
Question-9 In an AP the fourth and sixth term are 8 and 14 respectively find the
(i) first term
(ii) common difference
(iii) sum of first 20 term (2019)
Answer-9
T4 = 8 and T6 = 14
⇒ a + 3d = 8 … (i)
⇒ a + 5d = 14 … (ii)
Subtracting (i) from (ii),
(a + 5d) – (a + 3d) = 14 – 8
5d – 3d = 6
2d = 6
d = 6/2
d = 3
Hence, common difference d = 3
Substituting the value of d in (i),
a + 3(3) = 8
a = 8 – 9
a = -1
Hence, the first term = -1
(iii) The sum of first 20 terms
n = 20
S20 = n/2 × [2a + (n – 1)d]
= 20/2 × [2(-1) + (20 – 1)(3)]
= 10 × [-2 + (19)(3)]
= 10 × [-2 + 57]
= 10 × 55
= 550
Hence The sum of first 20 terms is 550
Question -10
(i) Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.
(ii) If the 4th term of an A.P. is 22 and 15th term is 66, find the first term Common difference Hence find the sum of 8 term of the AP (2018)
Answer -10
(i) Sum of first 51 terms of an A.P. in which
T2 = 14, T3 = 18
common difference d = T3 – T2
= 18 – 14, = 4
Where, a = T1 = 14 – 4 = 10
n = 51
We know that,
S51 = (n/2) (2a + (n – 1)d)
= (51/2) ((2 × 10) + (51 – 1)4)
= (51/2) (20 + (50 × 4))
= (51/2) (20 + 200)
= (51/2) × 220
= 5610
(ii) T4 = 22, T15 = 66
⇒ a + 3d = 22 … (i)
⇒ a + 14d = 66 … (ii)
Subtracting (i) from (ii), we get
(a + 14d) – (a + 3d) = 66 – 22
14d – 3d = 44
11d = 44
d = 44/11
d = 4
So, common difference d = 4
Substituting the value of d in (i), we have
a + 3(4) = 22
a = 22 – 12
a = 10
Hence, the first term a = 10
Now, the sum of first 8 terms of the A.P is
n = 8
Sn = n/2 × [2a + (n – 1)d]
S8 = 8/2 × [2(10) + (8 – 1)(4)]
= 10 × [20 + (7)(4)]
= 10 × [20 + 28]
= 10 × 48
= 480
ML Aggarwal AP GP Class 10 ICSE Maths Solutions
Arithmetic and Geometric Progression Exe-9.3
Question- 11 If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Answer- 11
S6 = 36
S16 = 256
Sn = (n/2) (2a + (n – 1)d)
S6 = (6/2) (2a + (6 – 1)d) = 36
3 (2a + 5d) = 36
Divide both the side by 3,
2a + 5d = 12 … equation (i)
Now, S16 = (16/2) (2a + (16 – 1)d) = 256
8 (2a + 15d) = 256
Divide both the side by 8,
2a + 15d = 32 … equation (ii)
Then, subtract equation (ii) from equation (i)
(2a + 5d) – (2a + 15d) = 12 – 32
2a + 5d – 2a – 15d = -20
-10d = -20
d = -20/-10
d = 2
now substitute the value of d in equation (i) to find a,
2a + 5d = 12
2a + 5(2) = 12
2a + 10 = 12
2a = 12 – 10
2a = 2
a = 1
So, S10 = (n/2) (2a + (n – 1)d)
= (10/2) ((2 × 1) + (10 – 1)2)
= 5 (2 + 18)
= 5 (20)
= 100
Hence, the sum of first 10 terms is 100
Question -12 Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
Answer -12
an = 3 + 4n
a1 = 3 + 4 x 1 = 3 + 4 = 7
a2 = 3 + 4 x 2 = 3 + 8 = 11
a3 = 3 + 4 x 3 = 3 + 12 = 15
a4 = 3 + 4 x 4 = 3 + 16 = 19
So, The numbers are 7, 11, 15, 19, ….
Then, first term a = 7, common difference d = 11 – 7 = 4
We know that,
S15 = (n/2) (2a + (n – 1)d)
= (15/2) ((2 × 7) + (15 – 1) × 4)
= (15/2) (14 + (14 × 4))
= (15/2) (14 + 56)
= (15/2) × 70
= 525
Hence, the sum of first 15 terms is 525.
Question -13 Sum of first six terms of an arithmetic progression is 42 and the ratio of 10th term and 30th term is 1: 3 calculate first and thirteen term .
Answer -13
S6 = 42 and T10/T30 = 1/3
We know that,
Sn = (n/2) (2a + (n – 1)d)
Tn = a + (n – 1)d
So, we have
S6 = (6/2) × [2a + (6 – 1)d] = 42
3 × (2a + 5d) = 42
2a + 5d = 14 … (i) Dividing by 3
And,
T10/T30 = [a + (10 – 1)d]/ [a + (30 – 1)d] = 1/3
(a + 9d)/ (a + 29d) = 1/3
On cross-multiplication,
3(a + 9d) = a + 29d
3a + 27d = a + 29d
3a – a + 27d – 29d = 0
2a – 2d = 0
a – d = 0
a = d
Using the above the relation in (i),
2a + 5a = 14
7a = 14
a = 14/7, a = 2
Hence, the first term of the A.P is 2.
the thirteenth term
T13 = 2 + (13 – 1)(2)
= 2 + 12 × 2
= 2 + 24, = 26
Question-14 In an AP the sum of first n term is 6n – n². Find is 25th term
Answer-14
Sn = 6n – n²
T25 = ?
S(n–1) = 6(n – 1) – (n – 1)2
= 6n – 6 – (n2 – 2n + 1)
= 6n – 6 – n2 + 2n –1
= 8n – n2 – 7
an = Sn – Sn – 1
= 6n – n2 – 8n + n2 + 7
= –2n + 7
a25 = –2(25) + 7
= –50 + 7
= –43.
Question-15 If Sn donate the sum of first n term of an AP prove that Sn = 3( S20 – S10 )
Answer-15
Sn = n/2 × [2a + (n – 1)d]
S10 = 10/2 × [2a + (10 – 1)d]
= 5 × (2a + 9d)
= 10a + 45d
S20 = 20/2 × [2a + (20 – 1)d]
= 10 × (2a + 19d)
= 20a + 190d
S30 = 30/2 × [2a + (30 – 1)d]
= 15 × (2a + 29d)
= 30a + 435d
taking L.H.S we have
3 (S20 – S10) = 3 [20a + 190d – (10a + 45d)]
= 3 (20a + 190d – 10a – 45d)
= 3 (10a + 145d)
= 30a + 435d
= S30
= R.H.S
– Hence proved
Arithmetic and Geometric Progression (AP GP )Exe-9.3
Question-16
(i) find the sum of first 1000 positive integers.
(ii) find the sum of first 15 multiple of 8.
Answer-16
(i) the sum of first 1000 positive integers.
Let 1 + 2 + 3 + …….. + 1000
This is an A.P with first term a = 1 and common difference d = 1
We know that,
Sn = n/2 × [2a + (n – 1)d]
S1000 = 1000/2 × [2(1) + (1000 – 1)(1)]
= 500 × (2 + 99)
= 500 × 101
= 50500
the sum of first 1000 positive integers is 50500.
(ii) The first 15 multiples of 8 are:
8, 16, 24, ….. where n = 15
This forms an A.P with first term a = 8 and common difference d = 16 – 8 = 8
Sn = n/2 × [2a + (n – 1)d]
S15 = 15/2 × [2(8) + (14 – 1)(8)]
= 15/2 × (16 + 13 × 8)
= 15/2 × (16 + 104)
= 15/2 × 120
= 15 × 60
= 900
Hence the sum of first 15 multiples of 8 is 900
Question-17
(i) Find the sum of two digit natural numbers which are divisible by 4
(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4
(iii) Find the sum of all multiple of 9 lying between 300 and 700
(iv) Find the sum of all natural numbers less than 100 which are divisible by 6
Answer-17
(i) The two-digit natural numbers which are divisible by 4 are:
4, 8, 12, 16, …..
This form an A.P.
The last term in this series is found out by dividing 100 by 4
100/4 = 25 and remainder is zero
So, the last two-digit number which is divisible by 4 is 96
And, it is the 24th term
So, we have
a = 4, d = 4, last term l = 96 and n = 24
Thus, sum of these numbers is
S24 = 24/2 (4 + 96)
= 12 × 100
= 1200
(ii) The natural numbers between 100 and 200 which are divisible by 4 are:
104, 108, 112, …..
This form an A.P.
The last term in this series is found out by dividing 200 by 4
200/4 = 50 and remainder is zero
So, the last natural number between 100 and 200 which is divisible by 4 is 196
So, we have
a = 104, d = 4 and last term l = 96
The number of terms is found out by
Tn = 196 = 104 + (n – 1)(4)
196 = 104 + 4n – 4
4n = 196 – 104 + 4
4n = 96
n = 96/4
n = 24
sum of these numbers is
S24 = 24/2 (104 + 196)
= 12 × 200
= 2400
(iii) The multiples of 9 lying between 300 and 700 are:
306, 315, 324, ……
This form an A.P. where a = 306 and d = 9
The last term in this series is found out by dividing 700 by 9
700/9 gives 77 as quotient and 7 as remainder.
the last number between 300 and 700 which is a multiple is 700 – 7 = 693
a = 306, d = 9 and l = 693
The number of terms
Tn = 693 = 306 + (n – 1)(9)
693 = 306 + 9n – 9
9n = 693 – 306 + 9
9n = 396
n = 396/9
n = 44
, sum of these numbers is
S44 = 44/2 (306 + 693)
= 22 × 999
= 21978
(iv) The natural numbers less than 100 which are divisible by 4 are:
4, 8, 12, …., 96
This forms an A.P. where a = 4, d = 4 and l = 96
Tn = 96 = 4 + (n – 1)4
96 = 4 + 4n – 4
4n = 96
n = 96/4
n = 24
sum of these numbers is
S24 = 24/2 (4 + 96)
= 12 × 100
= 1200
— : End of ML Aggarwal Arithmetic and Geometric Progression Exe-9.3 Class 10 ICSE Maths Solutions (AP GP) : –
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