A.P. 1/15, 1/12, 1/10, ... Here, a = 1/15 and d = 1/12 - (1/15) = 1/60. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d ⇒ S₁₁ = 11/2[2(1/15) + (11 - 1) (1/60)] ⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.
A.P. 1/15, 1/12, 1/10, …
Here, a = 1/15 and d = 1/12 – (1/15) = 1/60.
The sum of n terms of an AP is Given by
Sn = n/2[2a + (n -1)d]
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d
⇒ S₁₁ = 11/2[2(1/15) + (11 – 1) (1/60)]
⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.
A.P. : 0.6, 1.7, 2.8, ... Here, a = 0.6, and d = 1.7 - 0.6 = 1.1. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] ⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1) ⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505
A.P. : 0.6, 1.7, 2.8, …
Here, a = 0.6, and d = 1.7 – 0.6 = 1.1.
The sum of n terms of an AP is Given by
Sn = n/2[2a + (n -1)d]
⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1)
⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505
Each section of each class will plant tree = 3 × Class, therefore Total number of tree planted by class i = 3 × 1 = 3 Total number of tree planted by class ii = 3 × 2 = 6 Total number of tree planted by class iii = 3 × 3 = 9 Similary, the series of trees planted by classes are as follows: 3, 6, 9, .Read more
Each section of each class will plant tree = 3 × Class, therefore
Total number of tree planted by class i = 3 × 1 = 3
Total number of tree planted by class ii = 3 × 2 = 6
Total number of tree planted by class iii = 3 × 3 = 9
Similary, the series of trees planted by classes are as follows: 3, 6, 9, …., 36
Here, a = 3, d = 6 – 3 = 3 and n = 12.
The sum of n terms of an AP is given by S_n = n/2[2a + (n -1)d]
⇒ S₁₂ = 12 /2[2(3) + (12 – 1)(3)]
= 6[6 + 33] = 6(39) = 234
Hence, the total number of tree planted by the students is 234.
Here, a = - 5 and d = - 8 - (-5) = -3. Let, the nth term of the A.P. is - 230. Therefore, a_n = -230 ⇒ a + (n -1)d = - 230 ⇒ - 5 + (n - 1)(-3) = - 230 ⇒ (n - 1)(-3) = -225 ⇒ n - 1 = 75 ⇒ n = 76 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₇₆ = 76/2[- 5 - 230] ⇒ S₇₆ = 76/2[- 235] = - 38Read more
Here, a = – 5 and d = – 8 – (-5) = -3.
Let, the nth term of the A.P. is – 230.
Therefore, a_n = -230
⇒ a + (n -1)d = – 230
⇒ – 5 + (n – 1)(-3) = – 230
⇒ (n – 1)(-3) = -225
⇒ n – 1 = 75 ⇒ n = 76
The sum of n terms of an AP is given by
S_n = n/2[a +l]
⇒ S₇₆ = 76/2[- 5 – 230]
⇒ S₇₆ = 76/2[- 235]
= – 38 × 235 = – 8930
Here, a = 7 and d = 10(1/2) - 7 = 21/2 - 7 = 7/2. Let, the nth term of the A.P. is 84. Therefore, a_n = 84 ⇒ a + (n -1)d = 84 ⇒ 7 + (n - 1)(7/2) = 84 ⇒ (n - 1)(7/2) = 77 ⇒ n - 1 = 22 ⇒ n = 23 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₂₃ = 23/2[7 + 84] ⇒ S₂₃ = 23/2[91] = 2093/2 = 104Read more
Here, a = 7 and d = 10(1/2) – 7 = 21/2 – 7 = 7/2.
Let, the nth term of the A.P. is 84.
Therefore, a_n = 84
⇒ a + (n -1)d = 84
⇒ 7 + (n – 1)(7/2) = 84
⇒ (n – 1)(7/2) = 77
⇒ n – 1 = 22
⇒ n = 23
The sum of n terms of an AP is given by
S_n = n/2[a +l]
⇒ S₂₃ = 23/2[7 + 84]
⇒ S₂₃ = 23/2[91] = 2093/2 = 1046 (1/2)
Find the sum of the following APs: 1/15, 1/12, 1/10, …, to 11 terms.
A.P. 1/15, 1/12, 1/10, ... Here, a = 1/15 and d = 1/12 - (1/15) = 1/60. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d ⇒ S₁₁ = 11/2[2(1/15) + (11 - 1) (1/60)] ⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.
A.P. 1/15, 1/12, 1/10, …
See lessHere, a = 1/15 and d = 1/12 – (1/15) = 1/60.
The sum of n terms of an AP is Given by
Sn = n/2[2a + (n -1)d]
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d
⇒ S₁₁ = 11/2[2(1/15) + (11 – 1) (1/60)]
⇒ S₁₁ = 11/2[2/15+ 1/6] = 11/2[9/30] = 33/20.
Find the sum of the following APs: 0.6, 1.7, 2.8, . . ., to 100 terms.
A.P. : 0.6, 1.7, 2.8, ... Here, a = 0.6, and d = 1.7 - 0.6 = 1.1. The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] ⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1) ⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505
A.P. : 0.6, 1.7, 2.8, …
See lessHere, a = 0.6, and d = 1.7 – 0.6 = 1.1.
The sum of n terms of an AP is Given by
Sn = n/2[2a + (n -1)d]
⇒ S_100 = 100/2[2(0.6) + (100 -1)(1.1)
⇒ S_100 = 50[1.2 + 99 × 1.1] = 50[110.1] = 5505
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?
Each section of each class will plant tree = 3 × Class, therefore Total number of tree planted by class i = 3 × 1 = 3 Total number of tree planted by class ii = 3 × 2 = 6 Total number of tree planted by class iii = 3 × 3 = 9 Similary, the series of trees planted by classes are as follows: 3, 6, 9, .Read more
Each section of each class will plant tree = 3 × Class, therefore
See lessTotal number of tree planted by class i = 3 × 1 = 3
Total number of tree planted by class ii = 3 × 2 = 6
Total number of tree planted by class iii = 3 × 3 = 9
Similary, the series of trees planted by classes are as follows: 3, 6, 9, …., 36
Here, a = 3, d = 6 – 3 = 3 and n = 12.
The sum of n terms of an AP is given by S_n = n/2[2a + (n -1)d]
⇒ S₁₂ = 12 /2[2(3) + (12 – 1)(3)]
= 6[6 + 33] = 6(39) = 234
Hence, the total number of tree planted by the students is 234.
Find the sums given below: –5 + (–8) + (–11) + . . . + (–230)
Here, a = - 5 and d = - 8 - (-5) = -3. Let, the nth term of the A.P. is - 230. Therefore, a_n = -230 ⇒ a + (n -1)d = - 230 ⇒ - 5 + (n - 1)(-3) = - 230 ⇒ (n - 1)(-3) = -225 ⇒ n - 1 = 75 ⇒ n = 76 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₇₆ = 76/2[- 5 - 230] ⇒ S₇₆ = 76/2[- 235] = - 38Read more
Here, a = – 5 and d = – 8 – (-5) = -3.
See lessLet, the nth term of the A.P. is – 230.
Therefore, a_n = -230
⇒ a + (n -1)d = – 230
⇒ – 5 + (n – 1)(-3) = – 230
⇒ (n – 1)(-3) = -225
⇒ n – 1 = 75 ⇒ n = 76
The sum of n terms of an AP is given by
S_n = n/2[a +l]
⇒ S₇₆ = 76/2[- 5 – 230]
⇒ S₇₆ = 76/2[- 235]
= – 38 × 235 = – 8930
Find the sums given below : 7 + 10(1/2) + 14 + … + 84
Here, a = 7 and d = 10(1/2) - 7 = 21/2 - 7 = 7/2. Let, the nth term of the A.P. is 84. Therefore, a_n = 84 ⇒ a + (n -1)d = 84 ⇒ 7 + (n - 1)(7/2) = 84 ⇒ (n - 1)(7/2) = 77 ⇒ n - 1 = 22 ⇒ n = 23 The sum of n terms of an AP is given by S_n = n/2[a +l] ⇒ S₂₃ = 23/2[7 + 84] ⇒ S₂₃ = 23/2[91] = 2093/2 = 104Read more
Here, a = 7 and d = 10(1/2) – 7 = 21/2 – 7 = 7/2.
See lessLet, the nth term of the A.P. is 84.
Therefore, a_n = 84
⇒ a + (n -1)d = 84
⇒ 7 + (n – 1)(7/2) = 84
⇒ (n – 1)(7/2) = 77
⇒ n – 1 = 22
⇒ n = 23
The sum of n terms of an AP is given by
S_n = n/2[a +l]
⇒ S₂₃ = 23/2[7 + 84]
⇒ S₂₃ = 23/2[91] = 2093/2 = 1046 (1/2)