Here, a₃ = 12 and a_50 = 106. To find : a_29 Given that: a₃ = a + (3 - 1)d = 12 ⇒ a + 2d = 12 ⇒ a = 12 - 2d . . .(1) and a_50 = 106 ⇒ a + 49d = 106 Putting the value of a from equation (1), we get 12 - 2d + 49d = 106 ⇒ 47d = 94 ⇒ d = 2 putting the value of d in equation (1), we get a = 12 - 2(2) = 8Read more
Here, a₃ = 12 and a_50 = 106. To find : a_29
Given that: a₃ = a + (3 – 1)d = 12
⇒ a + 2d = 12
⇒ a = 12 – 2d . . .(1)
and a_50 = 106
⇒ a + 49d = 106
Putting the value of a from equation (1), we get
12 – 2d + 49d = 106
⇒ 47d = 94 ⇒ d = 2
putting the value of d in equation (1), we get
a = 12 – 2(2) = 8
Therefore, a₂₉ = a + 28d = 8 + 28(2) = 64
Hence, the 29th term of the AP is 64.
Let the first term of the first AP = A and common difference = d Let the first term of the second AP = a and the common difference = d Difference between their 100th term A_100 - a_100 ⇒ (A + 99d) - (a + 99d) = 100 ⇒ A - a = 100 Difference between their 1000th term = A_1000 - a_1000 = (A + 999d) - (Read more
Let the first term of the first AP = A and common difference = d
Let the first term of the second AP = a and the common difference = d
Difference between their 100th term A_100 – a_100
⇒ (A + 99d) – (a + 99d) = 100
⇒ A – a = 100
Difference between their 1000th term = A_1000 – a_1000
= (A + 999d) – (a + 999d)
= A – a = 100 [.: A – a = 100]
Hence, the difference between their 1000th term is 100.
Three digit numbers divisible by 7 : 105, 112, 119, . . . , 994 Let the total number of these numbers be n. Here, a = 105 and d = 112 - 105 = 7. To find: n, where a_n = 994. Given that: a_n = a + (n - 1)d = 994 ⇒ 105 + (n -1)(7) = 994 ⇒ 7(n -1) = 889 ⇒ n - 1 = 888/7 = 127 ⇒ n = 128 Hence, there areRead more
Three digit numbers divisible by 7 : 105, 112, 119, . . . , 994
Let the total number of these numbers be n.
Here, a = 105 and d = 112 – 105 = 7. To find: n, where a_n = 994.
Given that: a_n = a + (n – 1)d = 994
⇒ 105 + (n -1)(7) = 994
⇒ 7(n -1) = 889
⇒ n – 1 = 888/7 = 127 ⇒ n = 128
Hence, there are 128 three digits numbers which are divisible by 7.
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.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Here, a₃ = 12 and a_50 = 106. To find : a_29 Given that: a₃ = a + (3 - 1)d = 12 ⇒ a + 2d = 12 ⇒ a = 12 - 2d . . .(1) and a_50 = 106 ⇒ a + 49d = 106 Putting the value of a from equation (1), we get 12 - 2d + 49d = 106 ⇒ 47d = 94 ⇒ d = 2 putting the value of d in equation (1), we get a = 12 - 2(2) = 8Read more
Here, a₃ = 12 and a_50 = 106. To find : a_29
See lessGiven that: a₃ = a + (3 – 1)d = 12
⇒ a + 2d = 12
⇒ a = 12 – 2d . . .(1)
and a_50 = 106
⇒ a + 49d = 106
Putting the value of a from equation (1), we get
12 – 2d + 49d = 106
⇒ 47d = 94 ⇒ d = 2
putting the value of d in equation (1), we get
a = 12 – 2(2) = 8
Therefore, a₂₉ = a + 28d = 8 + 28(2) = 64
Hence, the 29th term of the AP is 64.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Let the first term of the first AP = A and common difference = d Let the first term of the second AP = a and the common difference = d Difference between their 100th term A_100 - a_100 ⇒ (A + 99d) - (a + 99d) = 100 ⇒ A - a = 100 Difference between their 1000th term = A_1000 - a_1000 = (A + 999d) - (Read more
Let the first term of the first AP = A and common difference = d
See lessLet the first term of the second AP = a and the common difference = d
Difference between their 100th term A_100 – a_100
⇒ (A + 99d) – (a + 99d) = 100
⇒ A – a = 100
Difference between their 1000th term = A_1000 – a_1000
= (A + 999d) – (a + 999d)
= A – a = 100 [.: A – a = 100]
Hence, the difference between their 1000th term is 100.
How many three-digit numbers are divisible by 7?
Three digit numbers divisible by 7 : 105, 112, 119, . . . , 994 Let the total number of these numbers be n. Here, a = 105 and d = 112 - 105 = 7. To find: n, where a_n = 994. Given that: a_n = a + (n - 1)d = 994 ⇒ 105 + (n -1)(7) = 994 ⇒ 7(n -1) = 889 ⇒ n - 1 = 888/7 = 127 ⇒ n = 128 Hence, there areRead more
Three digit numbers divisible by 7 : 105, 112, 119, . . . , 994
See lessLet the total number of these numbers be n.
Here, a = 105 and d = 112 – 105 = 7. To find: n, where a_n = 994.
Given that: a_n = a + (n – 1)d = 994
⇒ 105 + (n -1)(7) = 994
⇒ 7(n -1) = 889
⇒ n – 1 = 888/7 = 127 ⇒ n = 128
Hence, there are 128 three digits numbers which are divisible by 7.
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.