Let the first term of the AP = a and common difference = d Third term = 16 ⇒ a₃ = 16 ⇒ a + 2d = 16 ... (1) 7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12 ⇒ a + 6d = a + 4d + 12 ⇒ 2d = 12 ⇒ d = 6 Putting the value of d in equation (1), we get a + 2(6) = 16 ⇒ a = 4 Hence, the A.P = a, a +Read more
Let the first term of the AP = a and common difference = d
Third term = 16
⇒ a₃ = 16 ⇒ a + 2d = 16 … (1)
7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12 ⇒ d = 6
Putting the value of d in equation (1), we get a + 2(6) = 16
⇒ a = 4
Hence, the A.P = a, a + 2d, … = 4, 10, 16, …
Here, a = - 10, d = 7 - 10 = - 3 and n = 30. Therefore, putting the values in a_n = a + (n -1)d, we get a_30 = 10 +(30 -1) (- 3) ⇒ a_30 = 10 - 87 = - 77 Hence, the option (C) is correct.
Here, a = – 10, d = 7 – 10 = – 3 and n = 30.
Therefore, putting the values in a_n = a + (n -1)d, we get
a_30 = 10 +(30 -1) (- 3)
⇒ a_30 = 10 – 87 = – 77
Hence, the option (C) is correct.
Here, a = - 3, d = - 1/2 (- 3) = 5/2 and n = 11. Therefore, putting the values in a_n = a + (n -1)d, we get a₁₁ = - 3 +(11 -1) (5/2) ⇒ a₁₁ = - 3 + 25 = 22 Hence, the option (B) is correct.
Here, a = – 3, d = – 1/2 (- 3) = 5/2 and n = 11.
Therefore, putting the values in a_n = a + (n -1)d, we get
a₁₁ = – 3 +(11 -1) (5/2)
⇒ a₁₁ = – 3 + 25 = 22
Hence, the option (B) is correct.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Let the first term of the AP = a and common difference = d Third term = 16 ⇒ a₃ = 16 ⇒ a + 2d = 16 ... (1) 7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12 ⇒ a + 6d = a + 4d + 12 ⇒ 2d = 12 ⇒ d = 6 Putting the value of d in equation (1), we get a + 2(6) = 16 ⇒ a = 4 Hence, the A.P = a, a +Read more
Let the first term of the AP = a and common difference = d
See lessThird term = 16
⇒ a₃ = 16 ⇒ a + 2d = 16 … (1)
7th term exceeds the 5th term by 12, therefore a₇ = a₅ + 12
⇒ a + 6d = a + 4d + 12
⇒ 2d = 12 ⇒ d = 6
Putting the value of d in equation (1), we get a + 2(6) = 16
⇒ a = 4
Hence, the A.P = a, a + 2d, … = 4, 10, 16, …
Choose the correct choice in the following and justify : 30th term of the AP: 10, 7, 4, . . . , is
Here, a = - 10, d = 7 - 10 = - 3 and n = 30. Therefore, putting the values in a_n = a + (n -1)d, we get a_30 = 10 +(30 -1) (- 3) ⇒ a_30 = 10 - 87 = - 77 Hence, the option (C) is correct.
Here, a = – 10, d = 7 – 10 = – 3 and n = 30.
See lessTherefore, putting the values in a_n = a + (n -1)d, we get
a_30 = 10 +(30 -1) (- 3)
⇒ a_30 = 10 – 87 = – 77
Hence, the option (C) is correct.
Choose the correct choice in the following and justify : 11th term of the AP: – 3, – 1/2, 2, . . . , is
Here, a = - 3, d = - 1/2 (- 3) = 5/2 and n = 11. Therefore, putting the values in a_n = a + (n -1)d, we get a₁₁ = - 3 +(11 -1) (5/2) ⇒ a₁₁ = - 3 + 25 = 22 Hence, the option (B) is correct.
Here, a = – 3, d = – 1/2 (- 3) = 5/2 and n = 11.
See lessTherefore, putting the values in a_n = a + (n -1)d, we get
a₁₁ = – 3 +(11 -1) (5/2)
⇒ a₁₁ = – 3 + 25 = 22
Hence, the option (B) is correct.