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CBSE Class 11-science Answered

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Asked by Kchawla94 | 04 Mar, 2019, 12:43: PM
answered-by-expert Expert Answer
5 natural numbers that forms a sequence,  which are in Arithmetic progression and their sum is multiple of 15 is given by
 
n×(1+2+3+4+5) ,  where n is any positive integer so that their sum 15n which is multiple of 15.
 
Now let us answer the 5 questions.
 
(A) Atleast one of the number is multiple of 3  - True
         one of the number in the sequence will be 3n, multiple of 3
 
(B) Atleast two numbers are even - True
        there are two numbers 2n and 4n which are even irresoective of n is even or odd.
 
(C) Atleast two numbers are odd : False
       if n is even integer, all the numbers are even integer
 
(D) Sum of first and last term is multiple of 3 : True
         sum of first and last term is 6n which is multiple of 3 for any n
 
(E) Average of 5 numbers can not be multiple of 3 : False
        Average = 15n/5 = 3n . Hence Average is multiple of 3 for any n
Answered by Thiyagarajan K | 04 Mar, 2019, 04:17: PM

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