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.
The amount paid in the from of penalty is in the from of following AP: Rs 200, Rs 250, Rs 300, Rs 350, ... Here, a = 200, d = 250 - 200 = 50 and n = 30. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S_30 = 30/2[2(200) + (30 - 1)(50)] = 15[400 + 1450] = 15 (1850) = 27750 Hence, tRead more
The amount paid in the from of penalty is in the from of following AP:
Rs 200, Rs 250, Rs 300, Rs 350, …
Here, a = 200, d = 250 – 200 = 50 and n = 30.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S_30 = 30/2[2(200) + (30 – 1)(50)]
= 15[400 + 1450]
= 15 (1850) = 27750
Hence, the contractor has to pay Rs 27750 as penalty for the delay of 30 days.
Let the amount for first prize = Rs x number of prizes = 7 Total prize amount = Rs 700 Therefore, the series of 7 prizes are as follows: (x) + (x - 20) + (x - 40) + (x - 60) + (x - 80) + (x - 100) + (x - 120) = 700 ⇒ 7x - 420 = 700 ⇒ 7x = 1120 ⇒ x = 1120/7 = 160 Hence, the seven prizes are Rs 160, RRead more
Let the amount for first prize = Rs x
number of prizes = 7
Total prize amount = Rs 700
Therefore, the series of 7 prizes are as follows:
(x) + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = 700
⇒ 7x – 420 = 700
⇒ 7x = 1120
⇒ x = 1120/7 = 160
Hence, the seven prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40.
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.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: rupay 200 for the first day, rupay 250 for the second day, rupay 300 for the third day, etc., the penalty for each succeeding day being rupay 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
The amount paid in the from of penalty is in the from of following AP: Rs 200, Rs 250, Rs 300, Rs 350, ... Here, a = 200, d = 250 - 200 = 50 and n = 30. The sum of n terms of an AP is given by S_n = n/2[2a + (n - 1)d] ⇒ S_30 = 30/2[2(200) + (30 - 1)(50)] = 15[400 + 1450] = 15 (1850) = 27750 Hence, tRead more
The amount paid in the from of penalty is in the from of following AP:
See lessRs 200, Rs 250, Rs 300, Rs 350, …
Here, a = 200, d = 250 – 200 = 50 and n = 30.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n – 1)d]
⇒ S_30 = 30/2[2(200) + (30 – 1)(50)]
= 15[400 + 1450]
= 15 (1850) = 27750
Hence, the contractor has to pay Rs 27750 as penalty for the delay of 30 days.
A sum of rupay 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is rupay 20 less than its preceding prize, find the value of each of the prizes.
Let the amount for first prize = Rs x number of prizes = 7 Total prize amount = Rs 700 Therefore, the series of 7 prizes are as follows: (x) + (x - 20) + (x - 40) + (x - 60) + (x - 80) + (x - 100) + (x - 120) = 700 ⇒ 7x - 420 = 700 ⇒ 7x = 1120 ⇒ x = 1120/7 = 160 Hence, the seven prizes are Rs 160, RRead more
Let the amount for first prize = Rs x
See lessnumber of prizes = 7
Total prize amount = Rs 700
Therefore, the series of 7 prizes are as follows:
(x) + (x – 20) + (x – 40) + (x – 60) + (x – 80) + (x – 100) + (x – 120) = 700
⇒ 7x – 420 = 700
⇒ 7x = 1120
⇒ x = 1120/7 = 160
Hence, the seven prizes are Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60 and Rs 40.