Here, a₂ = 14, a₃ = 18 and n = 51. a_n = a + (n - 1)d ⇒ a₂ = a + (2 -1)d ⇒ 14 = a + d ⇒ a = 14 - d ...(1) and a₃ = a + (3 -1)d ⇒ 18 = a + 2d Putting the value of a from equation (1), we get ⇒ 18 = 14 - d + 2d ⇒ d = 4 Putting the value of d in equation (1), we get ⇒ a = 14 - 4 = 10 The sum of n termsRead more
Here, a₂ = 14, a₃ = 18 and n = 51.
a_n = a + (n – 1)d
⇒ a₂ = a + (2 -1)d
⇒ 14 = a + d
⇒ a = 14 – d …(1)
and a₃ = a + (3 -1)d
⇒ 18 = a + 2d
Putting the value of a from equation (1), we get
⇒ 18 = 14 – d + 2d
⇒ d = 4
Putting the value of d in equation (1), we get
⇒ a = 14 – 4 = 10
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S₅₁ = 51/2[2(10) +(51 -1) (4)]
⇒ S₅₁ = 51/2[220] = 5610.
Here, S₇ = 49 and S₁₇ = 289. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S₇ = 7/2[2a +(7 -1) d] ⇒ 49 = 7/2[2a + 6d] ⇒ 49 = 7(a + 3d) ⇒ 7 = a + 3d ⇒ a = 7 - 3d ...(1) and S₁₇ = 17/2[2a + 17 - 1)d) ⇒ 289 = 17/2[2a + 16d] ⇒ 289 = 17(a +8d) ⇒ 17 = a + 8d Putting the value of a froRead more
Here, S₇ = 49 and S₁₇ = 289.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S₇ = 7/2[2a +(7 -1) d]
⇒ 49 = 7/2[2a + 6d]
⇒ 49 = 7(a + 3d)
⇒ 7 = a + 3d
⇒ a = 7 – 3d …(1)
and S₁₇ = 17/2[2a + 17 – 1)d)
⇒ 289 = 17/2[2a + 16d]
⇒ 289 = 17(a +8d)
⇒ 17 = a + 8d
Putting the value of a from equation (1), we get
⇒ 17 = 7 – 3d + 8d
⇒ 5d = 10
⇒ d = 2
Putting the value of d in equation (1), we get
⇒ a = 7 – 3 × 2 = 1
The sum of n terms of AP is given by
S_n = n/2[2a + (n – 1)d]
= n/2[2(1) + (n -1)(2)]
= n/2[2 + 2n – 2] = n²
The first 40 positive integers divisible by 6 are 6, 12, 18, ..., 240. Here, a = 6, d = 12 - 6 = 6 and n = 40. The sum of n terms of an AP is Given by S_n = n/2[2a + (n -1)d] ⇒ S_40 = 10/2[2(6) + (40 -1)(6)] = 20[12 + 234] = 20(246) = 4920 Hence, the sum of the first 40 positive integers divisible bRead more
The first 40 positive integers divisible by 6 are 6, 12, 18, …, 240.
Here, a = 6, d = 12 – 6 = 6 and n = 40.
The sum of n terms of an AP is Given by
S_n = n/2[2a + (n -1)d]
⇒ S_40 = 10/2[2(6) + (40 -1)(6)]
= 20[12 + 234]
= 20(246) = 4920
Hence, the sum of the first 40 positive integers divisible by 6 is 4920.
The first 15 multiples of 8 are 8, 16, 24, ..., 120. Here, a = 8,d = 16 - 8 = 8 and n = 15. The sum of n terms of an AP is gevin by S_n = n/2[2a + (n -1)d] ⇒ S₁₅ = 15/2[2(8) + (15 -1)(8)] = 15/2[16 + 112] = 15/2(128) = 960 Hence, the sum of the 15 multiples of is 960.
The first 15 multiples of 8 are 8, 16, 24, …, 120.
Here, a = 8,d = 16 – 8 = 8 and n = 15.
The sum of n terms of an AP is gevin by
S_n = n/2[2a + (n -1)d]
⇒ S₁₅ = 15/2[2(8) + (15 -1)(8)]
= 15/2[16 + 112]
= 15/2(128)
= 960
Hence, the sum of the 15 multiples of is 960.
The odd numbers between 0 and 50: 1, 3, 5, ..., 49. Here, a = 1, d = 3 - 1 = 2 and n = 25. The sum of n terms of an Ap is given by S_n = n/2[2a + (n -1)d] ⇒ S₂₅ = 25/2[2(1) + (25 - 1)(2)] = 25/2[2 + 48] = 25/2(50) = 625 Hence, the sum of the odd number between 0 and 50 is 625.
The odd numbers between 0 and 50: 1, 3, 5, …, 49.
Here, a = 1, d = 3 – 1 = 2 and n = 25.
The sum of n terms of an Ap is given by
S_n = n/2[2a + (n -1)d]
⇒ S₂₅ = 25/2[2(1) + (25 – 1)(2)]
= 25/2[2 + 48]
= 25/2(50) = 625
Hence, the sum of the odd number between 0 and 50 is 625.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Here, a₂ = 14, a₃ = 18 and n = 51. a_n = a + (n - 1)d ⇒ a₂ = a + (2 -1)d ⇒ 14 = a + d ⇒ a = 14 - d ...(1) and a₃ = a + (3 -1)d ⇒ 18 = a + 2d Putting the value of a from equation (1), we get ⇒ 18 = 14 - d + 2d ⇒ d = 4 Putting the value of d in equation (1), we get ⇒ a = 14 - 4 = 10 The sum of n termsRead more
Here, a₂ = 14, a₃ = 18 and n = 51.
See lessa_n = a + (n – 1)d
⇒ a₂ = a + (2 -1)d
⇒ 14 = a + d
⇒ a = 14 – d …(1)
and a₃ = a + (3 -1)d
⇒ 18 = a + 2d
Putting the value of a from equation (1), we get
⇒ 18 = 14 – d + 2d
⇒ d = 4
Putting the value of d in equation (1), we get
⇒ a = 14 – 4 = 10
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S₅₁ = 51/2[2(10) +(51 -1) (4)]
⇒ S₅₁ = 51/2[220] = 5610.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Here, S₇ = 49 and S₁₇ = 289. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S₇ = 7/2[2a +(7 -1) d] ⇒ 49 = 7/2[2a + 6d] ⇒ 49 = 7(a + 3d) ⇒ 7 = a + 3d ⇒ a = 7 - 3d ...(1) and S₁₇ = 17/2[2a + 17 - 1)d) ⇒ 289 = 17/2[2a + 16d] ⇒ 289 = 17(a +8d) ⇒ 17 = a + 8d Putting the value of a froRead more
Here, S₇ = 49 and S₁₇ = 289.
See lessThe sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S₇ = 7/2[2a +(7 -1) d]
⇒ 49 = 7/2[2a + 6d]
⇒ 49 = 7(a + 3d)
⇒ 7 = a + 3d
⇒ a = 7 – 3d …(1)
and S₁₇ = 17/2[2a + 17 – 1)d)
⇒ 289 = 17/2[2a + 16d]
⇒ 289 = 17(a +8d)
⇒ 17 = a + 8d
Putting the value of a from equation (1), we get
⇒ 17 = 7 – 3d + 8d
⇒ 5d = 10
⇒ d = 2
Putting the value of d in equation (1), we get
⇒ a = 7 – 3 × 2 = 1
The sum of n terms of AP is given by
S_n = n/2[2a + (n – 1)d]
= n/2[2(1) + (n -1)(2)]
= n/2[2 + 2n – 2] = n²
Find the sum of the first 40 positive integers divisible by 6.
The first 40 positive integers divisible by 6 are 6, 12, 18, ..., 240. Here, a = 6, d = 12 - 6 = 6 and n = 40. The sum of n terms of an AP is Given by S_n = n/2[2a + (n -1)d] ⇒ S_40 = 10/2[2(6) + (40 -1)(6)] = 20[12 + 234] = 20(246) = 4920 Hence, the sum of the first 40 positive integers divisible bRead more
The first 40 positive integers divisible by 6 are 6, 12, 18, …, 240.
See lessHere, a = 6, d = 12 – 6 = 6 and n = 40.
The sum of n terms of an AP is Given by
S_n = n/2[2a + (n -1)d]
⇒ S_40 = 10/2[2(6) + (40 -1)(6)]
= 20[12 + 234]
= 20(246) = 4920
Hence, the sum of the first 40 positive integers divisible by 6 is 4920.
Find the sum of the first 15 multiples of 8.
The first 15 multiples of 8 are 8, 16, 24, ..., 120. Here, a = 8,d = 16 - 8 = 8 and n = 15. The sum of n terms of an AP is gevin by S_n = n/2[2a + (n -1)d] ⇒ S₁₅ = 15/2[2(8) + (15 -1)(8)] = 15/2[16 + 112] = 15/2(128) = 960 Hence, the sum of the 15 multiples of is 960.
The first 15 multiples of 8 are 8, 16, 24, …, 120.
See lessHere, a = 8,d = 16 – 8 = 8 and n = 15.
The sum of n terms of an AP is gevin by
S_n = n/2[2a + (n -1)d]
⇒ S₁₅ = 15/2[2(8) + (15 -1)(8)]
= 15/2[16 + 112]
= 15/2(128)
= 960
Hence, the sum of the 15 multiples of is 960.
Find the sum of the odd numbers between 0 and 50.
The odd numbers between 0 and 50: 1, 3, 5, ..., 49. Here, a = 1, d = 3 - 1 = 2 and n = 25. The sum of n terms of an Ap is given by S_n = n/2[2a + (n -1)d] ⇒ S₂₅ = 25/2[2(1) + (25 - 1)(2)] = 25/2[2 + 48] = 25/2(50) = 625 Hence, the sum of the odd number between 0 and 50 is 625.
The odd numbers between 0 and 50: 1, 3, 5, …, 49.
See lessHere, a = 1, d = 3 – 1 = 2 and n = 25.
The sum of n terms of an Ap is given by
S_n = n/2[2a + (n -1)d]
⇒ S₂₅ = 25/2[2(1) + (25 – 1)(2)]
= 25/2[2 + 48]
= 25/2(50) = 625
Hence, the sum of the odd number between 0 and 50 is 625.