Find the sum of all multiples of 5 between 100 and 300 inclusive.
Ans: 8200
I do not know how to do this. Really do appreciate step by step workings and explanation.
Thanks!
100 , 105, 110, ... 290 , 295 ,300
That is arithmetic progression.
The initial term of this arithmetic progression is a1 = 100
The common difference of successive members is d = 5
The n-th term this arithmetic progression an = 300
Use formula for n-th term of the sequence :
an = a1 + ( n - 1 ) * d
In this case :
a1 = 100
an = 300
d = 5
an = a1 + ( n - 1 ) * d
300 = 100 + ( n - 1 ) * 5 Divide both sides by 5
60 = 20 + ( n - 1 )
60 = 20 + n - 1
60 = 19 + n Subtract 19 to both sides
60 - 19 = 19 + n - 19
41 = n
n = 41
The sum of the members of a arithmetic progression is :
Sn = ( n / 2 ) * ( a1 + an ) = n * ( a1 + an ) / 2
Sn = 41 * ( 100 + 300 ) / 2
Sn = 41 * 400 / 2
Sn = 16,400 / 2
Sn = 8,200
Well, let's break it down step by step, shall we?
First, let's find the first multiple of 5 within the range. We divide 100 by 5, and we get 20. So, the first multiple of 5 is 20 x 5 = 100.
Next, let's find the last multiple of 5 within the range. We divide 300 by 5, and we get 60. So, the last multiple of 5 is 60 x 5 = 300.
Now, let's find the number of terms in this sequence. To do that, we subtract the first multiple from the last multiple, and then add 1. So, the number of terms is (300 - 100) + 1 = 201.
Now, let's find the sum of this arithmetic sequence. To do that, we use the formula:
Sum = (n/2) x (first term + last term)
where n is the number of terms. So, the sum of all the multiples of 5 between 100 and 300 is:
Sum = (201/2) x (100 + 300)
= 201/2 x 400
= 402 x 100
= 40200.
Oops! I made a miscalculation there, let me correct it! My bad!
Actually, the sum of all the multiples of 5 between 100 and 300 is:
Sum = (n/2) x (first term + last term)
= (201/2) x (100 + 300)
= (201/2) x 400
= 40200
So, the sum of all the multiples of 5 between 100 and 300 is 40200.
Hope that helps! If you have any more questions, feel free to ask!
To find the sum of all multiples of 5 between 100 and 300 inclusive, you can use the formula for the sum of an arithmetic series.
Step 1: Identify the first term (a) and the common difference (d).
In this case, the first term (a) is 100 and the common difference (d) is 5.
Step 2: Identify the last term (L).
The last term (L) is the largest multiple of 5 that is less than or equal to 300. In this case, it is 300.
Step 3: Determine the number of terms (n).
To find the number of terms, you can use the formula: n = (L - a) / d + 1.
Substituting the values, we get: n = (300 - 100) / 5 + 1 = 201 / 5 + 1 = 40 + 1 = 41.
Step 4: Compute the sum using the formula: S = (n / 2) * (a + L).
Substituting the values, we get: S = (41 / 2) * (100 + 300) = 20.5 * 400 = 8200.
Therefore, the sum of all multiples of 5 between 100 and 300 inclusive is 8200.
To find the sum of all multiples of 5 between 100 and 300 inclusive, you can follow these steps:
Step 1: Determine the first multiple of 5 within the given range.
In this case, the first multiple of 5 within the range is 100.
Step 2: Determine the last multiple of 5 within the given range.
In this case, the last multiple of 5 within the range is 300.
Step 3: Find the number of terms (n) in the arithmetic sequence of multiples of 5.
To find the number of terms, you can use the formula: n = (last term – first term) / common difference + 1.
In this case, the first term is 100, the last term is 300, and the common difference is 5.
So, n = (300 - 100) / 5 + 1 = 41.
Step 4: Use the formula for the sum of an arithmetic sequence to find the sum.
The sum of an arithmetic sequence can be calculated using the formula: S = (n/2)(first term + last term).
In this case, n = 41, the first term is 100, and the last term is 300.
So, S = (41/2)(100 + 300) = 20.5 * 400 = 8200.
Therefore, the sum of all multiples of 5 between 100 and 300 inclusive is 8200.
300 = 100 + 40*5
So, set things up as an arithmetic sequence, and find that
S41 = 41/2 (100+300) = 8200