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.
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.
Let the first term = a and common difference = d According to question, a₁₇ = a_10 + 7 ⇒ a + 16d = a + 9d + 7 ⇒ 7d = 7 ⇒ d = 1 Hence, the common difference is 1.
Let the first term = a and common difference = d
According to question, a₁₇ = a_10 + 7
⇒ a + 16d = a + 9d + 7
⇒ 7d = 7 ⇒ d = 1
Hence, the common difference is 1.
Here, a₃ = 4 and a₉ = - 8. To find : n, where a_n = 0 Given that: a₃ = a + (3 - 1)d = 4 ⇒ a + 2d = 4 ⇒ a = 4 - 2d . . .(1) and a₉ = - 8 ⇒ a + 8d = - 8 Putting the value of a from equation (1), we get 4 - 2d + 8d = -8 ⇒ 6d = - 12 ⇒ d = -2 putting the value of d in equation (1), we get a = 4 - 2(- 2)Read more
Here, a₃ = 4 and a₉ = – 8. To find : n, where a_n = 0
Given that: a₃ = a + (3 – 1)d = 4
⇒ a + 2d = 4
⇒ a = 4 – 2d . . .(1)
and a₉ = – 8
⇒ a + 8d = – 8
Putting the value of a from equation (1), we get
4 – 2d + 8d = -8
⇒ 6d = – 12 ⇒ d = -2
putting the value of d in equation (1), we get
a = 4 – 2(- 2) = 8
putting the values in a_n = 0, we get
a_n = a + (n -1)d = 0
⇒ 8 +(n -1)(- 2) = 0
⇒ n – 1 = 1 = 4 ⇒ n = 5
Hence, the 5th term of this AP is zero.
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.
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.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Let the first term = a and common difference = d According to question, a₁₇ = a_10 + 7 ⇒ a + 16d = a + 9d + 7 ⇒ 7d = 7 ⇒ d = 1 Hence, the common difference is 1.
Let the first term = a and common difference = d
See lessAccording to question, a₁₇ = a_10 + 7
⇒ a + 16d = a + 9d + 7
⇒ 7d = 7 ⇒ d = 1
Hence, the common difference is 1.
If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Here, a₃ = 4 and a₉ = - 8. To find : n, where a_n = 0 Given that: a₃ = a + (3 - 1)d = 4 ⇒ a + 2d = 4 ⇒ a = 4 - 2d . . .(1) and a₉ = - 8 ⇒ a + 8d = - 8 Putting the value of a from equation (1), we get 4 - 2d + 8d = -8 ⇒ 6d = - 12 ⇒ d = -2 putting the value of d in equation (1), we get a = 4 - 2(- 2)Read more
Here, a₃ = 4 and a₉ = – 8. To find : n, where a_n = 0
See lessGiven that: a₃ = a + (3 – 1)d = 4
⇒ a + 2d = 4
⇒ a = 4 – 2d . . .(1)
and a₉ = – 8
⇒ a + 8d = – 8
Putting the value of a from equation (1), we get
4 – 2d + 8d = -8
⇒ 6d = – 12 ⇒ d = -2
putting the value of d in equation (1), we get
a = 4 – 2(- 2) = 8
putting the values in a_n = 0, we get
a_n = a + (n -1)d = 0
⇒ 8 +(n -1)(- 2) = 0
⇒ n – 1 = 1 = 4 ⇒ n = 5
Hence, the 5th term of this AP is zero.