Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-12) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 5.51

Q.12. Find the sum of
(i) the first 15 multiples of 8
(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.
(iii) all 3 − digit natural numbers which are divisible by 13.
(iv) all 3 − digit natural numbers, which are multiples of 11.
(v) all 2 − digit natural numbers divisible by 4.
(vi) first 8 multiples of 3.
Ans. In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
S_n = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
(i) First 15 multiples of 8.
So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.
Also, all these terms will form an A.P. with the common difference of 8.
So here,
First term (a) = 8
Number of terms (n) = 15
Common difference (d) = 8
Now, using the formula for the sum of n terms, we get
S_n = 15/2[2(8) + (15 - 1)8]
= 15/2[16 + (14)8]
= 15/2(16 + 112)
= (15/2)(128) = 960
Therefore, the sum of the first 15 multiples of 8 is 960.
(ii) (a) First 40 positive integers divisible by 3
So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.
Also, all these terms will form an A.P. with the common difference of 3.
So here,
First term (a) = 3
Number of terms (n) = 40
Common difference (d) = 3
Now, using the formula for the sum of n terms, we get
S_n = 40/2[2(3) + (40 - 1)3]
= 20[6 + (39)3]
= 20(6 + 117)
= 20(123) = 2460
Therefore, the sum of first 40 multiples of 3 is 2460.
(b) First 40 positive integers divisible by 5
So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.
Also, all these terms will form an A.P. with the common difference of 5.
So here,
First term (a) = 5
Number of terms (n) = 40
Common difference (d) = 5
Now, using the formula for the sum of n terms, we get
S_n = 40/2[2(5) + (40 - 1)5]
= 20[10 + (39)5]
= 20(10 + 195)
= 20(205) = 4100
Therefore, the sum of first 40 multiples of 3 is 4100
(c) First 40 positive integers divisible by 6
So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.
Also, all these terms will form an A.P. with the common difference of 6.
So here,
First term (a) = 6
Number of terms (n) = 40
Common difference (d) = 6
Now, using the formula for the sum of n terms, we get
S_n = 40/2[2(6) + (40 - 1)6]
= 20[12 + (39)6]
= 20(12 + 234)
= 20(246) = 4920
Therefore, the sum of first 40 multiples of 3 is 4920.
(iii) All 3 digit natural number which are divisible by 13
So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.
Also, all these terms will form an A.P. with the common difference of 13.
So here,
First term (a) = 104
Last term (l) = 988
Common difference (d) = 13
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
988 = 104 + (n - 1)13
988 = 104 + 13n - 13
988 = 91 + 13n
Further simplifying,

n = 897/13 = 69
Now, using the formula for the sum of n terms, we get

= 69/2 (208 + 884)
On further simplification, we get,
S_n = (69/2)(1092)
= 69(546) = 37674
Therefore, the sum of all the 3 digit multiples of 13 is S_n = 37674.
(iv) all 3-digit natural numbers, which are multiples of 11.
We know that the first 3 digit number multiple of 11 will be 110.
Last 3 digit number multiple of 11 will be 990.
So here,
First term (a) = 110
Last term (l) = 990
Common difference (d) = 11
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
990 = 110 + (n − 1)11
990 = 110 + 11n − 11
990 = 99 + 11n
891 = 11n
81 = n
Now, using the formula for the sum of n terms, we get
S_n = 81/2[2(110) + (81 − 1)11]
S_n = 81/2[220 + 80 × 11]
S_n = (81/2) × 1100
S_n = 81 × 550
S_n = 44550
Therefore, the sum of all the 3 digit multiples of 11 is 44550.
(v) 2-digit no. divisible by 4 are 12,16,20,........,96
We can see it forms an AP as the common difference is 4 and the first term is 4.
To find no. of terms n,
we know that
a_n = a + (n - 1)d
96 = 12 + (n − 1)4
84 = (n − 1)4
21 = n − 1
22 = n
Now,
First term (a) = 12
Number of terms (n) = 22
Common difference (d) = 4
Now, using the formula for the sum of n terms, we get
S₂₂ = 22/2{2(12) + (22 − 1)4}
S₂₂ = 11{24 + 84}
S₂₂ = 1188
Hence, the sum of 22 terms is 1188 which are divisible by 4.
(vi) First 8 multiples of 3 are { 3, 6, 9...,24}
We can observe they are in AP with first term (a) = 3 and last term (l) = 24 and number of terms are 8.

S₈ = 4×(3 + 24) = 108
Hence, the sum of the first 8 multiples of 3 is 108.

Ques.13. Find the sum:
(i) 2 + 4 + 6 ... + 200
(ii) 3 + 11 + 19 + ... + 803
(iii) (−5) + (−8)+ (−11) + ... + (−230)
(iv) 1 + 3 + 5 + 7 + ... + 199
(v) 7 + 10(1/2) + 14 +...+ 84
(vi) 34 + 32 + 30 + ... + 10
(vii) 25 + 28 + 31 + ... + 100
(viii) 18 + 15(1/2) + 13 +...+ (−49(1/2))
Ans. In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of n terms of an A.P.,
S_n = n/2[2a + (n - 1)d]
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
(i) 2 + 4 + 6 ... + 200
Common difference of the A.P. (d) = a₂ - a₁
= 6 - 4 = 2
So here,
First term (a) = 2
Last term (l) = 200
Common difference (d) = 2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
200 = 2 + (n - 1)2
200 = 2 + 2n - 2
200 = 2n
Further simplifying,
n = 200/2 = 100
Now, using the formula for the sum of n terms, we get
S_n = 100/2[2(2) + (100 - 1)2]
= 50[4 + (99)2]
= 50(4 + 198)
On further simplification, we get,
S_n = 50(202) = 10100
Therefore, the sum of the A.P is S_n = 10100
(ii) 3 + 11 + 19 + ... + 803
Common difference of the A.P. (d) = a₂ - a₁
= 19 - 11 = 8
So here,
First term (a) = 3
Last term (l) = 803
Common difference (d) = 8
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
Further simplifying,
803 = 3 + (n - 1)8
803 = 3 + 8n - 8
803 + 5  = 8n
808 = 8n
n = 808/8 = 101
Now, using the formula for the sum of n terms, we get

= 101/2[6 + (100)8]
= 101/2(806)
= 101(403) = 40703
Therefore, the sum of the A.P is S_n = 40703
(iii) (-5) + (-8) + (-11) + ... + (-230)
Common difference of the A.P. (d) = a₂ - a₁
= -8 - (-5)
= -8 + 5 = -3
So here,
First term (a) = −5
Last term (l) = −230
Common difference (d) = −3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
-230 = -5 + (n - 1)(-3)
-230 = -5 - 3n + 3
-230 + 2 = -3n
-228/-3 = n
n = 76
Now, using the formula for the sum of n terms, we get
S_n = 76/2[2(-5) + (76 - 1)(-3)]
= 38[-10 + (75) (-3)]
= 38(-10 - 225)
= 38(-235) = -8930
Therefore, the sum of the A.P is S_n = -8930
(iv) 1 + 3 + 5 + 7 ... + 199
Common difference of the A.P. (d) = a₂ - a₁
= 3 - 1 = 2
So here,
First term (a) = 1
Last term (l) = 199
Common difference (d) = 2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
199 = 1 + (n - 1)2
199 = 1 + 2n - 2
199 + 1 = 2n
n = 200/2 = 100
Now, using the formula for the sum of n terms, we get
S_n = 100/2[2(1) + (100 - 1)2]
= 50[2 + (99)2]
= 50(2 + 198)
On further simplification, we get,
S_n = 50(200) = 10000
Therefore, the sum of the A.P is S_n = 10000
(v) 7 + 10(1/2) + 14 + ... + 84
Common difference of the A.P is
d = 10(1/2) - 7
= (21/2) - 7

= 7/2
So here,
First term (a) = 7
Last term (l) = 84
Common difference (d) = 7/2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
84 = 7 + (n - 1)(7/2)

84(2) = 7 + 7n
Further solving for n,
7n = 168 - 7
n = 161/7 = 23
Now, using the formula for the sum of n terms, we get

= (23/2)(91)
On further simplification, we get,
S_n = 2093/2
Therefore, the sum of the A.P is S_n = 2093/2
(vi) 34 + 32 + 30 + ... + 10
Common difference of the A.P. (d) = a₂ - a₁
= 32 - 34 = -2
So here,
First term (a) = 34
Last term (l) = 10
Common difference (d) = −2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,
10 = 34 + (n - 1)(-2)
10 = 34 - 2n + 2
10 = 36 - 2n
10 - 36 = -2n
Further solving for n,
-2n = -26
n = -26/-2
n = 13
Now, using the formula for the sum of n terms, we get
S_n = 13/2[2(34) + (13 - 1)(-2)]
= 13/2[68 + (12)(-2)]
= 13/2(68 - 24)
= (13/2)(44)
On further simplification, we get,
S_n = 13(22) = 286
Therefore, the sum of the A.P is S_n = 286
(vii) 25 + 28 + 31 + ... + 100
Common difference of the A.P. (d) = a₂ - a₁
= 28 - 25 = 3
So here,
First term (a) = 25
Last term (l) = 100
Common difference (d) = 3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1 )d
So, for the last term,
100 = 25 + (n - 1)(3)
100 = 25 + 3n - 3
100 = 22 + 3n
100 - 22 = 3n
Further solving for n,
78 = 3n
n = 78/3 = 26
Now, using the formula for the sum of n terms, we get
S_n = 26/2 [2(25) + (26 - 1)(3)]
= 13[50 + (25)(3)]
= 13(50 + 75)
= 13(125)
On further simplification, we get,
S_n = 1625
Therefore, the sum of the A.P is S_n = 1625.
(viii) 18 + 15(1/2) + 13 + ... + ((-49)(1/2)).
Common difference of the A.P. (d) = a2 - a1
= 15(1/2) - 18
= 31/2 - 18

So here,
First term (a) = 18
Last term (l) =
Common difference (d) = (-5)/2
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
a_n = a + (n - 1)d
So, for the last term,

n = 28
Now, using the formula for the sum of n terms, we get

S_n = -441
Therefore, the sum of the A.P is S_n = -441.