Multiples of 4 lie between 10 and 250: 12, 16, 20, ..., 248 Let the total number of multiples of 4 lie between 10 and 250 be n. Here, a = 12 and d = 16 - 12 = 4. To find: n, where a_n = 248. Given that: a_n = a + (n - 1)d = 248 ⇒ 12 + (n -1)(4) = 248 ⇒ 4(n -1) = 236 ⇒ n - 1 = 236/4 = 59 ⇒ n = 60 HenRead more
Multiples of 4 lie between 10 and 250: 12, 16, 20, …, 248
Let the total number of multiples of 4 lie between 10 and 250 be n.
Here, a = 12 and d = 16 – 12 = 4. To find: n, where a_n = 248.
Given that: a_n = a + (n – 1)d = 248
⇒ 12 + (n -1)(4) = 248 ⇒ 4(n -1) = 236
⇒ n – 1 = 236/4 = 59 ⇒ n = 60
Hence, the total number of multiples of 4 lie between 10 and 250 is 60.
First term of first AP = A = 63 and common difference = D = 65 - 63 = 2 Therefore, A_n = A + (n -1)D ⇒ A_n = 63 + (n -1)2 First term of second AP = a = 3 and common difference = d = 10 - 3 = 7 Therefore, a_n = a + (n -1)d ⇒ a_n = 3 +(n -1)7 According to question, A_n = a_n ⇒ 63 + (n - 1)2 = 3 + (n -Read more
First term of first AP = A = 63 and common difference = D = 65 – 63 = 2
Therefore, A_n = A + (n -1)D ⇒ A_n = 63 + (n -1)2
First term of second AP = a = 3 and common difference = d = 10 – 3 = 7
Therefore, a_n = a + (n -1)d ⇒ a_n = 3 +(n -1)7
According to question, A_n = a_n
⇒ 63 + (n – 1)2 = 3 + (n -1)7 ⇒ 63 + 2_n – 2 = 3 + 7_n – 7
⇒ 65 = 5_n ⇒ n = 13
Hence, the 13th term of both the APs are equal.
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.
How many multiples of 4 lie between 10 and 250?
Multiples of 4 lie between 10 and 250: 12, 16, 20, ..., 248 Let the total number of multiples of 4 lie between 10 and 250 be n. Here, a = 12 and d = 16 - 12 = 4. To find: n, where a_n = 248. Given that: a_n = a + (n - 1)d = 248 ⇒ 12 + (n -1)(4) = 248 ⇒ 4(n -1) = 236 ⇒ n - 1 = 236/4 = 59 ⇒ n = 60 HenRead more
Multiples of 4 lie between 10 and 250: 12, 16, 20, …, 248
See lessLet the total number of multiples of 4 lie between 10 and 250 be n.
Here, a = 12 and d = 16 – 12 = 4. To find: n, where a_n = 248.
Given that: a_n = a + (n – 1)d = 248
⇒ 12 + (n -1)(4) = 248 ⇒ 4(n -1) = 236
⇒ n – 1 = 236/4 = 59 ⇒ n = 60
Hence, the total number of multiples of 4 lie between 10 and 250 is 60.
For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
First term of first AP = A = 63 and common difference = D = 65 - 63 = 2 Therefore, A_n = A + (n -1)D ⇒ A_n = 63 + (n -1)2 First term of second AP = a = 3 and common difference = d = 10 - 3 = 7 Therefore, a_n = a + (n -1)d ⇒ a_n = 3 +(n -1)7 According to question, A_n = a_n ⇒ 63 + (n - 1)2 = 3 + (n -Read more
First term of first AP = A = 63 and common difference = D = 65 – 63 = 2
See lessTherefore, A_n = A + (n -1)D ⇒ A_n = 63 + (n -1)2
First term of second AP = a = 3 and common difference = d = 10 – 3 = 7
Therefore, a_n = a + (n -1)d ⇒ a_n = 3 +(n -1)7
According to question, A_n = a_n
⇒ 63 + (n – 1)2 = 3 + (n -1)7 ⇒ 63 + 2_n – 2 = 3 + 7_n – 7
⇒ 65 = 5_n ⇒ n = 13
Hence, the 13th term of both the APs are equal.
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.