Question

Five distinct positive integers are in a arithmetic progression with a positive common difference. If their sum is 10020, then the smallest possible value of the last term is
[1]
(a) 2002
(b) 2004
(C) 2006
(d) 2007​

Answer 1

Answer:

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Answer 2

The smallest possible value of last term=2006

Step-by-step explanation:

Let a be the first term and d be the common positive difference .

Then,A.P is

$a,a+d,a+2d,a+3d,a+4d$

Sum of five distinct positive integers=10020

Sum of  nth term of A.P

$S_n=\frac{n}{2}(a+a_n)$

Where a=First term

$a_n$=nth term of A.P

n=Total number of terms in A.P

Using the formula

$10020=\frac{5}{2}(a+a_n)$

$a+a_n=\frac{10020\times 2}{5}=4008$

nth term of A.P is given by

$a_n=a+(n-1)d$

Where d=Common difference of A.P

Using the formula

$a_5=a+4d$

Substitute the value

$a+a+4d=4008$

$2(a+2d)=4008$

$a+2d=\frac{4008}{2}=2004$

$a=2004-2d$

Substitute the value of a

$a_5=2004-2d+4d=2004+2d$

d=0 cannot be consider because d is positive

Therefore, possible value of d=1,2,3,..

Substitute d=1

$a_5=2004+2(1)=2006$

for d=2

$a_5=2004+2(2)=2008$

Hence, the smallest possible value of last term=2006

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