Starting salary = a = Rs 5000 annual increment (common difference) = d = Rs 200 Let, after n years, his salary become Rs 7000. Therefore, a_n = 7000 ⇒ a + (n -1)d = 7000 ⇒ 5000 + (n - 1)(200) = 7000 ⇒ (n -1)(200) = 2000 ⇒ n - 1 = 10 ⇒ n = 11 Hence, in 11th year his salary become Rs 7000.
Starting salary = a = Rs 5000 annual increment (common difference) = d = Rs 200
Let, after n years, his salary become Rs 7000.
Therefore, a_n = 7000
⇒ a + (n -1)d = 7000
⇒ 5000 + (n – 1)(200) = 7000
⇒ (n -1)(200) = 2000
⇒ n – 1 = 10 ⇒ n = 11
Hence, in 11th year his salary become Rs 7000.
Let the first term of the AP = a and common difference = d According to first condition, a₄ + a₈ = 24 ⇒ a + 3d + a + 7d = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ⇒ a = 12 - 5d ...(1) According to second condition, a₆ + a_10 = 44 ⇒ a + 5d + a + 9d = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 Putting the value of a frRead more
Let the first term of the AP = a and common difference = d
According to first condition, a₄ + a₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24 ⇒ a + 5d = 12
⇒ a = 12 – 5d …(1)
According to second condition, a₆ + a_10 = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22
Putting the value of a from equation (1), we get
(12 -5d) +7d = 22
⇒ 2d = 10 ⇒ d = 5
Putting the value of d in equation (1), we get a = 12 – 5(5) = – 13
The first three term of this AP: a, a + d, a + 2d = – 13, – 8, – 3.
Savings for the first week = a = Rs. 5 increment in saving = d = Rs. 1.75 Let, her saving become Rs. 20.75 after n weeks. Therefore, a_n = 20.75 ⇒ a + (n -1)d = 20.75 ⇒ 5 + (n -1)(1.75) = 20.75 ⇒ (n -1)(1.75) = 15.75 ⇒ n - 1 = 15.75/1.75 = 9 ⇒ n = 10 Hence, her saving become Rs 20.75, after 10 weeksRead more
Savings for the first week = a = Rs. 5 increment in saving = d = Rs. 1.75
Let, her saving become Rs. 20.75 after n weeks.
Therefore, a_n = 20.75
⇒ a + (n -1)d = 20.75
⇒ 5 + (n -1)(1.75) = 20.75
⇒ (n -1)(1.75) = 15.75
⇒ n – 1 = 15.75/1.75 = 9 ⇒ n = 10
Hence, her saving become Rs 20.75, after 10 weeks.
The 20th term from the last term of an AP: 3, 8, 13, ... 253 = the 20th term from the beginning of the AP: 253. 13, 8, 3. In the A.P.: 253, ..., 13, 8, 3, first term = 253 and common difference = 3 - 8 = -5 Therefore, a_20 = a + 19d ⇒ a_20 = 253 + 19(-5) = 253 - 95 = 158
The 20th term from the last term of an AP: 3, 8, 13, … 253 = the 20th term from the beginning of the AP: 253. 13, 8, 3.
In the A.P.: 253, …, 13, 8, 3, first term = 253 and common difference = 3 – 8 = -5
Therefore, a_20 = a + 19d
⇒ a_20 = 253 + 19(-5) = 253 – 95 = 158
First term = 3 and common difference = 15 - 3 = 12 Let the nth term of AP: 3, 15, 27, 39 ... will be 132 more than its 54th term. Therefore, a_n = a₅₄ + 132 ⇒ a + (n -1)d = a + 53d + 132 ⇒ (n -1)(12) = 53 × 12 + 132 ⇒ (n -1)(12) = 768 ⇒ n -1 = 768/12 = 64 ⇒ n = 65 Hence, 65th term of the AP: 3, 15,Read more
First term = 3 and common difference = 15 – 3 = 12
Let the nth term of AP: 3, 15, 27, 39 … will be 132 more than its 54th term.
Therefore, a_n = a₅₄ + 132
⇒ a + (n -1)d = a + 53d + 132
⇒ (n -1)(12) = 53 × 12 + 132
⇒ (n -1)(12) = 768
⇒ n -1 = 768/12 = 64
⇒ n = 65
Hence, 65th term of the AP: 3, 15, 27, 39 . . . will be 132 more than its 54th trem.
Subba Rao started work in 1995 at an annual salary of rupay 5000 and received an increment of rupay 200 each year. In which year did his income reach ` 7000?
Starting salary = a = Rs 5000 annual increment (common difference) = d = Rs 200 Let, after n years, his salary become Rs 7000. Therefore, a_n = 7000 ⇒ a + (n -1)d = 7000 ⇒ 5000 + (n - 1)(200) = 7000 ⇒ (n -1)(200) = 2000 ⇒ n - 1 = 10 ⇒ n = 11 Hence, in 11th year his salary become Rs 7000.
Starting salary = a = Rs 5000 annual increment (common difference) = d = Rs 200
See lessLet, after n years, his salary become Rs 7000.
Therefore, a_n = 7000
⇒ a + (n -1)d = 7000
⇒ 5000 + (n – 1)(200) = 7000
⇒ (n -1)(200) = 2000
⇒ n – 1 = 10 ⇒ n = 11
Hence, in 11th year his salary become Rs 7000.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Let the first term of the AP = a and common difference = d According to first condition, a₄ + a₈ = 24 ⇒ a + 3d + a + 7d = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ⇒ a = 12 - 5d ...(1) According to second condition, a₆ + a_10 = 44 ⇒ a + 5d + a + 9d = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 Putting the value of a frRead more
Let the first term of the AP = a and common difference = d
See lessAccording to first condition, a₄ + a₈ = 24
⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24 ⇒ a + 5d = 12
⇒ a = 12 – 5d …(1)
According to second condition, a₆ + a_10 = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44 ⇒ a + 7d = 22
Putting the value of a from equation (1), we get
(12 -5d) +7d = 22
⇒ 2d = 10 ⇒ d = 5
Putting the value of d in equation (1), we get a = 12 – 5(5) = – 13
The first three term of this AP: a, a + d, a + 2d = – 13, – 8, – 3.
Ramkali saved rupay 5 in the first week of a year and then increased her weekly savings by rupay 1.75. If in the nth week, her weekly savings become rupay 20.75, find n.
Savings for the first week = a = Rs. 5 increment in saving = d = Rs. 1.75 Let, her saving become Rs. 20.75 after n weeks. Therefore, a_n = 20.75 ⇒ a + (n -1)d = 20.75 ⇒ 5 + (n -1)(1.75) = 20.75 ⇒ (n -1)(1.75) = 15.75 ⇒ n - 1 = 15.75/1.75 = 9 ⇒ n = 10 Hence, her saving become Rs 20.75, after 10 weeksRead more
Savings for the first week = a = Rs. 5 increment in saving = d = Rs. 1.75
See lessLet, her saving become Rs. 20.75 after n weeks.
Therefore, a_n = 20.75
⇒ a + (n -1)d = 20.75
⇒ 5 + (n -1)(1.75) = 20.75
⇒ (n -1)(1.75) = 15.75
⇒ n – 1 = 15.75/1.75 = 9 ⇒ n = 10
Hence, her saving become Rs 20.75, after 10 weeks.
Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
The 20th term from the last term of an AP: 3, 8, 13, ... 253 = the 20th term from the beginning of the AP: 253. 13, 8, 3. In the A.P.: 253, ..., 13, 8, 3, first term = 253 and common difference = 3 - 8 = -5 Therefore, a_20 = a + 19d ⇒ a_20 = 253 + 19(-5) = 253 - 95 = 158
The 20th term from the last term of an AP: 3, 8, 13, … 253 = the 20th term from the beginning of the AP: 253. 13, 8, 3.
See lessIn the A.P.: 253, …, 13, 8, 3, first term = 253 and common difference = 3 – 8 = -5
Therefore, a_20 = a + 19d
⇒ a_20 = 253 + 19(-5) = 253 – 95 = 158
Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
First term = 3 and common difference = 15 - 3 = 12 Let the nth term of AP: 3, 15, 27, 39 ... will be 132 more than its 54th term. Therefore, a_n = a₅₄ + 132 ⇒ a + (n -1)d = a + 53d + 132 ⇒ (n -1)(12) = 53 × 12 + 132 ⇒ (n -1)(12) = 768 ⇒ n -1 = 768/12 = 64 ⇒ n = 65 Hence, 65th term of the AP: 3, 15,Read more
First term = 3 and common difference = 15 – 3 = 12
See lessLet the nth term of AP: 3, 15, 27, 39 … will be 132 more than its 54th term.
Therefore, a_n = a₅₄ + 132
⇒ a + (n -1)d = a + 53d + 132
⇒ (n -1)(12) = 53 × 12 + 132
⇒ (n -1)(12) = 768
⇒ n -1 = 768/12 = 64
⇒ n = 65
Hence, 65th term of the AP: 3, 15, 27, 39 . . . will be 132 more than its 54th trem.