Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 5.

Question

ScanSolve AI Tutor · Accepted Answer

To find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 5, we first need to identify the arithmetic sequence.

Identify the first term ( $a_1$ ).
The natural numbers between 250 and 1000 mean numbers strictly greater than 250 and strictly less than 1000.
The first number greater than 250 that is divisible by 5 is 255.
So, $a_1 = 255$ .
Identify the last term ( $a_n$ ).
The last number less than 1000 that is divisible by 5 is 995.
So, $a_n = 995$ .
Identify the common difference ( $d$ ).
Since the numbers are exactly divisible by 5, the common difference is 5.
So, $d = 5$ .
Find the number of terms ( $n$ ) in the sequence.
We use the formula for the $n^{th}$ term of an arithmetic sequence: $a_n = a_1 + (n-1)d$ .
Substitute the values:
$995 = 255 + (n-1)5$
Subtract 255 from both sides:
$995 - 255 = (n-1)5$ $740 = (n-1)5$
Divide by 5:
$\frac{740}{5} = n-1$ $148 = n-1$
Add 1 to both sides:
$n = 148 + 1$ $n = 149$
There are 149 terms in the sequence.
Find the sum of the terms ( $S_n$ ).
We use the formula for the sum of an arithmetic sequence: $S_n = \frac{n}{2}(a_1 + a_n)$ .
Substitute the values:
$S_{149} = \frac{149}{2}(255 + 995)$ $S_{149} = \frac{149}{2}(1250)$ $S_{149} = 149 \times 625$ $S_{149} = 93125$
The sum of all natural numbers between 250 and 1000 which are exactly divisible by 5 is $\boxed{\text{93125}}$ .

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