CBSE Solutions for Class 10 English

જવાબ :

જવાબ : d = a₂ – a₁ = 
  =  = -3

જવાબ : d = a₂ – a₁ = 
=  = -2

જવાબ : d = a₂ – a₁= 
= 

જવાબ : d = a₂ – a₁ = 
=  = -b

જવાબ : a₂₁ – a₇ = 84 …[Given in the question
∴ (a + 20d) – (a + 6d) = 84 …[a_n = a + (n – 1)d
20d – 6d = 84
14d = 84 ⇒ d  = 6

જવાબ : Here, a = 5, d = 9 – 5 = 4, l = 185
9^th term from the end = 185 – (9 – 1)4
= 185 – 8 × 4 = 185 – 32 = 153

જવાબ : 40°, 60°, 80°

જવાબ : no

જવાબ : 22nd term

જવાબ : 28^th term

જવાબ : Yes Let 1^st term = a, Common difference = d
a4 = 0 a + 3d = 0 ⇒ a = -3d … (A)
To prove: a₂₅ = 3 × a11
a + 24d = 3(a + 10d) …[From (A)
⇒ -3d + 24d = 3(-3d + 10d)
⇒ 21d = 21d
From above, a₂₅ = 3(a₁₁) (Hence proved)

જવાબ : 8, 10, 12,14 Solution:
Let ‘a’ be the 1st term and ‘d’ be the common difference.

જવાબ : Yes

Common difference d = 9 – 4
= 14 – 9 = 5
Given that the last term, l = 254, n = 10
nth term from the end = l – (n – 1) d
∴ 10th term from the end = 254 – (10 – 1) × 5
= 254 – 45 = 209

જવાબ : 6
Solution:
12, 18, 24, …,96
Here,First term a = 12, common diff. d = 18 – 12 = 6, an = 96
a + (n – 1)d = an
∴ 12 + (n – 1)6 = 96
⇒ (n − 1)6 = 96 – 12 = 84
⇒ n – 1 =  = 14
⇒ n = 14 + 1 = 15
∴ There are 15 two-digit numbers divisible by 6.

જવાબ : 43
Solution:
203, 210, 217, …, 497
Here, First term a = 203, common diff. d = 210 – 203 = 7, an = 497
∴ a + (n – 1) d = an
203 + (n – 1) 7 = 497
(n – 1) 7 = 497 – 203 = 294
n – 1 =  = 42 ∴ n = 42 + 1 = 43
∴ There are 43 natural nos. between 200 and 500 which are divisible by 7.

જવાબ : 30
Solution:
Two-digit numbers divisible by 3 are:
12, 15, 18, …, 99
Here, First term a = 12, common diff. d = 15 – 12 = 3, an = 99
∴ a + (n – 1)d = an
12 + (n – 1) (3) = 99
(n – 1) (3) = 99 – 12 = 87
n – 1 =  = 29
∴ n = 29 + 1 = 30
∴ There are 30 such numbers.

જવાબ : Yes
Solution:
“3 digits nos.” are 100, 101, 102, …, 999
3 digits nos. “divisible by 7” are:
105, 112, 119, 126, …, 994
first term a = 105, common diff. d = 7, an = 994, n = ?
As a + (n – 1)d = 994 = an
∴ 105 + (n – 1)7 = 994
(n − 1)7 = 994 – 105
(n – 1) =  = 127
∴ n = 127 + 1 = 128

જવાબ : Yes
Solution:
To find: Number of terms of A.P., i.e., n.
A.P. = 108 + 117 + 126 + … + 999
a = 108
d = 117 – 108 = 9
a_n = 999
a + (n – 1)d = an
∴ 108 + (n – 1) 9 = 999
⇒ (n − 1) 9 = 999 – 108 = 891
⇒ (n − 1) =  = 99
∴ n = 99 + 1 = 100

જવાબ : 89
Solution:
Numbers divisible by both 2 and 5 are 110, 120, 130, …, 990.
Here first term a = 110, common diff. d = 120 – 110 = 10, an = 990
As a + (n – 1)d = an = 990
110 + (n – 1)(10) = 990
(n – 1)(10) = 990 – 110 = 880
(n − 1) =  = 88
∴ n = 88 + 1 = 89

જવાબ : Yes

જવાબ : True

જવાબ : True

જવાબ : False

જવાબ : True

જવાબ : n=7

જવાબ : 65^th

જવાબ : 44^th

જવાબ : 6, 2, -2, -6 …….

જવાબ : d=3

જવાબ : n=38 & S_n =6973

જવાબ : n=6 & d=5

જવાબ : T₂₅=S₂₅−S₂₄T₂₅=S₂₅−S₂₄ = [3(25)² /2] +[13×25]/2−3(24)²/2 + [13×24]/2= 80

જવાબ : The AP formed is 3, 5, 7, 9, .........

જવાબ : middle most terms will 8 and 9
T8=−11+(8−1)4=17T8=−11+(8−1)4= 17
T9=−11+(9−1)4=21T9=−11+(9−1)4= 21

જવાબ : 11(11x−6y)/x+y

જવાબ : 1

જવાબ : True

જવાબ : False

જવાબ : Yes, d=5

જવાબ : Yes, d=0

જવાબ : No

જવાબ : Yes, d=11

જવાબ : Yes, d=-3

જવાબ : Next 2 terms are -11, -13

જવાબ : 3

જવાબ : 60

જવાબ : 6n -7

જવાબ : 10000

જવાબ : 40

જવાબ : Let A be the first term and D be the common difference of the given A.P.
p^th term = A + (p – 1)D = a …(A)
q^th term = A + (q – 1)D = b …(B)
r^th term = A + (r – 1)D = c … (C)
L.H.S. = (a – b)r + (b – c)p + (c – a)q
= [A + (p – 1)D – (A + (q – 1)D)]r + [A + (q – 1)D – (A + (r – 1)D)]p + [A + (r – 1)D – (A + (p – 1)D)]q
= [(p – 1 – q + 1)D]r + [(q – 1 – r + 1)D]p + [(r – 1 – p + 1)D]q
= D[(p – q)r + (q – r)p + (r – p)q]
= D[pr – qr + qp – rp + rq – pq]
= D[0] = 0 = R.H.S.

જવાબ :
From (i),
a = 2(4) – 5 = 8 – 5 = 3
As a_n = a + (n – 1) d
∴ a_n = 3 + (n – 1) 4 = 3 + 4n – 4
a_n = (4n – 1)

જવાબ :
From (i),
a = 2(4) – 3
= 8 – 3 = 5
n^th term = a + (n – 1) d
∴ n^th term = 5 + (n – 1) 4
= 5 + 4n – 4 = (4n + 1)

જવાબ :
From (A) and (B), a = 4 – 1 = 3
As n^th term = a + (n – 1) d
∴ n^th term = 3 + (n – 1) 4
= 3 + 4n -4 = 4n -1

જવાબ : a = 8, d = a₂ – a₁ = 10 – 8 = 2, n = 60
a₆₀ = a + 59d = 8 + 59(2) = 126
∴ Sum of its last 10 terms = S₆₀ – S₅₀
= (a + a_n) – (2a + (n – 1)d)
=  (8 + a₆₀) –  (2 × 8 + (50 – 1)2)
= 30 (8 + 126) – 25 (16 + 98)
= 4020 – 25 × 114
= 4020 – 2850 = 1170

જવાબ : Let, ‘a’ be the first term and ‘d’ be the common difference of A.P. respectively,
a = 5, d = 12 – 5 = 7, n = 50
∴ a_n = a + (n – 1)d
a₅₀ = 5 + 49(7) = 5 + 343 = 348
∴ Sum of its last 15 terms = S₅₀ – S₃₅
= (a + a_n) –  (2a + (n − 1)d)
=  (5 + 348) –  [2(5) + (35 – 1)7]
= 25(353) –  (10 + 238)
= 8825 – 35 × 124
= 8825 – 4340 = 4485

જવાબ : S_n = [2a + (n − 1) d] …(A)

Putting the value of d = 2 in (B), we get a = 7
∴ S_n = [2(7) + (n – 1). 2]
= .2 [7 + n – 1]
= n (n + 6) or n² + 6n

જવાબ : Here, Sum of 10 prizes = 1600

S₁₀ = 1600, d = -20, n = 10
S_n =  (2a + (n – 1)d]
∴ [2a + (10 – 1)(-20)] = 1600
2a – 180 = 320
2a = 320 + 180 = 500
a = 250
Hence, 1^st prize = a = ₹250
2^nd prize = a₂ = a + d = 250 + (-20) = ₹230
3^rd prize = a₃ = a₂ + d = 230 – 20 = ₹210
4^th prize = a₄ = a₃ + d = 210 – 20 = ₹190
5^th prize = a₅ = a₄ + d = 190 – 20 = ₹170
6^th prize = a₆ = a₅ + d = 170 – 20 = ₹150
7^th prize = a₇ = a₆ + d = 150 – 20 = ₹130
8^th prize = a₈ = a₇ + d = 130 – 20 = ₹110
9^th prize = a₉ = a₈ + d = 110 – 20 = ₹590
10^th prize = a₁₀ = a₉ + d = 90 – 20 = ₹70
= ₹ 1,600

જવાબ : Here a = 8, l = 350, d = 9
As we know, a + (n − 1) d = a₂
∴ 8 + (n – 1) 9 = 350
(n − 1) 9 = 350 – 8 = 342
n – 1 =  = 38
n = 38 + 1 = 39
∴ There are 39 terms.
∴ S_n = (a + l)
∴ S₃₉ =  (8 + 350) =  × 358
= 39 × 179 = 6981

જવાબ : Given: a = 7, S₂₀ = -240
Here, S_n = [2a + (n – 1)d]
∴ S₂₀ = [2(7) + (20 – 1)d]
-240 = 10(14 + 19d)
– = 14 + 19d = -24
19d = -24 – 14 = -38
⇒ d =  = -2 …(A)
Again, a_n = a + (n – 1)d
∴ a₂₄ = 7 + (24 – 1) (-2) … [From (A)
= 7 – 46 = -39

જવાબ : Here a = 5 … (A)
Here, a₁ + a₂ + a₃ + a₄ =  (a₅ + a₆ + a₇ + a₈)
∴ a + (a + d) + (a + 2d) + (a + 3d)
=  [(a + 4d) + (a + 5d) + (a + 6d) + (a + 780] …[a_n = a(n – 1)d
∴ 4a + 6d =  (4a + 22d)
8a + 12d = 4a + 22d
8a – 4a = 22d – 12d
4a = 100 ⇒ 4(5) = 10d
d =  = 2 …[From (A)
∴ Common difference, d = 2

જવાબ : As S_n = [2a + (n − 1)d]

S₂ = 119
From (A) and (B), a = 17 – 3(5) = 17 – 15 = 2
∴S_n = [2(2) + (n – 1)5]
= [4 + 5n – 5]
=  (5n – 1)

જવાબ : Solution:
a₂₄ = 2 (10) … [Given
a + 23d = 2 (a +9d) [∵ a_n = a + (n – 1)d)
23d = 2a + 18d – a
23d – 18d = a ⇒ a = 5d …(A)
To prove: a₇₂ = 4 (a₁₅)

From (ii) and (iii), L.H.S = R.H.S …Hence Proved

જવાબ : Given that, a = -12, d=-9 + 12 = 3
Here, a_n = a + (n – 1)d = 21
∴ -12 + (n – 1).3 = 21
(n – 1).3 = 21 + 12 = 33
∴ n – 1 = 11
Total number of terms,
n = 11 + 1 = 12
New A.P. is

Here 1st term, a = -11
Common difference, d = -8 + 11 = 3;
Last term, a_n = 22; Number of terms, n= 12

જવાબ : Here, n = 50,
Here, S₁₀ = 210
=  (2a + 9d) = 210 ….[S_n =  [2a+(n – 1)2]
5(2a + 9d) = 210
2a + 9d =  = 42
⇒ 2a = 42 – 9d ⇒ a =  …(A)
Now, 50 = (1 + 2 + 3 + …) + (36 + 37 + … + 50) Sum = 2565
Sum of its last 15 terms = 2565 …[Given
S₅₀ – S₃₅ = 2565
(2a + 49d) –  (2a + 34d) = 2565
100a + 2450d – 70a – 1190d = 2565 × 2
30a + 1260d = 5130
3a + 1260 = 513 …[Dividing both sides by 10

જવાબ :

જવાબ : Classes: 1 + I + II + … + XII
Sections: 2(I) + 2(II) + 2(III) + … + 2(XII)
Total no. of trees
= 2 + 4 + 6 … + 24
= (2 × 2) + (2 × 4) + (2 × 6) + … + (2 × 24)
= 4 + 8 + 12 + … + 48
:: S₁₂ = (4 + 48) = 6(52) = 312 trees

જવાબ : Money required by Ramkali for admission of her daughter = ₹2500
A.P. formed by saving
100, 120, 140, … upto 12 terms ….(A)
Let, a, d and n be the first term, common difference and number of terms respectively. Here, a = 100, d = 20, n = 12
S_n =  (2a + (n − 1)d)
⇒ S₁₂ =  (2(100) + (12 – 1)20)
S₁₂ =  [2(100) + 11(20)] = 6[420] = ₹2520

જવાબ :

જવાબ : We have, S_m = 4m² – m
Put m = 1,
S₁ = 4(1)² – (1)
= 4 – 1 = 3
Put m = 2,
S₂ = 4(2)² – 2
= 16 – 2 = 14
∴ a₁ = S₁ = 3 = a
a₂ = S₂ – S₁ = 14 – 3 = 11
d = a₂ – a₁
d = 11 – 3 = 8
a_n = a + (n – 1)d = 107 …[Given in the ques.
∴ 3+ (n – 1)8 = 107
(n – 1)8 = 107 – 3 = 104
(n – 1) =  = 13
n = 13 + 1 = 14
∴ a₂₁ = a + 20d
= 3 + (20)8
= 3 + 160 = 163

જવાબ : We have, S_q = 63q – 3q²
Put q = 1 in the equation
S₇ = 63(1) – 3(1)²
= 63 – 3 = 60
Put q = 2 in the equation
S2 = 63(2) – 3(2)²
= 126 – 12 = 114
∴ a = a₁ = S₁ = 60
a₂ = S₂ – S7₁ = 114 – 60 = 54
d = a₂ – a₁ = 54 – 60 = -6
p^th term = -60
a + (p – 1)d = -60
60 + (p – 1)(-6) = -60
(p – 1)(-6) = -60 – 60 = -120
(p – 1) =  = 20
p= 20 + 1 = 21
∴ a₁₁ = a + 10d
= 60 + 10(-6)
= 60 – 60 = 0

જવાબ :

1-C, 2-D, 3-A, 4-B

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-D, 2-A, 3-B, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-D, 2-A, 3-B, 4-C

જવાબ :

1-C, 2-D, 3-A, 4-B

જવાબ :

1-D, 2-C, 3-B, 4-A

જવાબ :

1-B, 2-A, 3-D, 4-C