Average of the number from 100 to 400 which are divisble by 13
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104 to 390 are divisible by 13
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Ans.
The first number and the last number divisible by 13 between 100 and 400 are 104 and 390 respectively.
Let us consider a AP series :
104 + 117 + 130 + ... + 390
First term = 104
and
Last term = 390
We are having common difference as 13, since the numbers are divisible by 13.
So, Common Difference = 13
We apply the formula for last term of a AP sequence :
Last term = First term + {(n-1l)×Common Difference}, where n is the number of terms in the series.
=> 390 = 104 + {(n-1)×13}
=> (n-1)×13 = 286
=> n - 1 = 22
=> n = 23
Therefore, there are 23 terms in the AP series.
Then the sum of the numbers (between 100 and 400), divisible by 13 is
= (Numbers of terms/2)×(First term + Last term)
= (23/2)×(104 + 390)
= 5681.
♧♧♧ FORMULA GUIDE ♧♧♧
Let, a and l are first and last term of any AP series containing n terms, and d is the common difference.
Then,
1. Last term (l) = a + (n-1)d
2. Sum of the series (S)
= (n/2)×(a + l)
= (n/2)×{2a + (n-1)d}
⬆HOPE THIS HELPS YOU⬅
Ans.
The first number and the last number divisible by 13 between 100 and 400 are 104 and 390 respectively.
Let us consider a AP series :
104 + 117 + 130 + ... + 390
First term = 104
and
Last term = 390
We are having common difference as 13, since the numbers are divisible by 13.
So, Common Difference = 13
We apply the formula for last term of a AP sequence :
Last term = First term + {(n-1l)×Common Difference}, where n is the number of terms in the series.
=> 390 = 104 + {(n-1)×13}
=> (n-1)×13 = 286
=> n - 1 = 22
=> n = 23
Therefore, there are 23 terms in the AP series.
Then the sum of the numbers (between 100 and 400), divisible by 13 is
= (Numbers of terms/2)×(First term + Last term)
= (23/2)×(104 + 390)
= 5681.
♧♧♧ FORMULA GUIDE ♧♧♧
Let, a and l are first and last term of any AP series containing n terms, and d is the common difference.
Then,
1. Last term (l) = a + (n-1)d
2. Sum of the series (S)
= (n/2)×(a + l)
= (n/2)×{2a + (n-1)d}
⬆HOPE THIS HELPS YOU⬅
Draxillus:
gr10 answer bhai
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