Question
Download Solution PDFFind the sum of all natural numbers lying between 100 and 1000, which are multiples of 5?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
Let us suppose a be the first term and d be the common difference of an AP. Then the nth term of an AP is given by:an = a + (n - 1) × d.
Note: If l is the last term of a sequence, then l = an = a + (n - 1) × d.
The sum of n terms of an AP with first term a and common difference d is given by:
Where l is the last term of the AP.
CALCULATION:
All the natural numbers lying between 100 and 1000 which are multiples of 5 is represented by the sequence given below:
105, 110, ......, 995
For the given sequence we have a = 105, d = 5 and l = 995
As we know that, if l is the last term of a sequence, then l = an = a + (n - 1) × d.
⇒ 995 = 105 + 5(n - 1)
⇒ n = 179
Thus, a = 105, l = 995 and n = 179
As we know that,
So, for the given sequence the sum of the sequence is given by:
Hence, the required sum is 98450
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