Question 1209312
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In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence, 
how many terms are multiples of 5?
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<pre>
The n-th term of this sequence,  {{{a[n]}}},  is the sum of the first n natural numbers
    {{{a[1]}}} = 1,

    {{{a[2]}}} = 3 = 1 + 2,

    {{{a[3]}}} = 6 = 1 + 2 + 3,

    {{{a[4]}}} = 10 = 1 + 2 + 3 + 4,

    {{{a[5]}}} = 15 = 1 + 2 + 3 + 4 + 5,

and so on.  For the sum of the first natural numbers, there is the formula

    {{{a[n]}}} = {{{(n(n+1))/2}}}.


So, the question is: how many terms  {{{a[n]}}}  of the first 1000 terms are multiples of 5.


In order for the term  {{{a[n]}}}  be a multiple of 5, EITHER  n  OR  (n+1)  must be multiple of 5
(here "either - or"  is exclusive).



Of the first 1000 numbers n = 1, 2, 3, 4, 5, . . . , 1000,  there are  1000/5 = 200 multiple of 5.

Of the first 1000 numbers n+1 = 2, 3, 4, 5, . . . , 1001,  there are  {{{(1001-1)/5}}} = 200 other numbers multiple of 5.



Thus, in all,  among first 1000 numbers  {{{a[n]}}} = {{{(n*(n+1))/2}}} ,  there are  200 + 200 = 400 numbers multiple of 5.


<U>ANSWER</U>.  In the sequence 1, 3, 6, 10, 15, 21, ..., in the first 1000 terms of the sequence, 
         400 numbers are multiple of 5.
</pre>

Solved.