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Q.

If  a 1 , a 2 , a 3 ,..., a 4001   are terms of an A.P. such that  1 a 1 a 2 + 1 a 2 a 3 +...+ 1 a 4000 a 4001 =10   and a 2 + a 4000 =50  , then find the value of  a 1 a 4001  .


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a

20

b

30

c

40

d

50 

answer is B.

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Detailed Solution

Given that the series a 1 , a 2 ,..., a 4001   are in A.P. From the series, the first terms is t 1 = a 1  . Let the common difference be d  .
So, d= a 2 a 1 =...= a 4001 a 4000  .
The general term of the series is given by,
t n = t 1 +(n1)d = a 1 +d(n1)  
Then,
a 2 = a 1 +(21)d= a 1 +d a 4000 = a 1 +(40001)d= a 1 +3999d a 4001 = a 1 +(40011)d= a 1 +4000d  
Given that, a 2 + a 4000 =50  .
a 2 + a 4000 =50 a 1 +d + a 1 +3999d =50 a 1 + a 1 +4000d=50  
a 1 + a 4001 =50  … (1)
Also, given that, 1 a 1 a 2 + 1 a 2 a 3 +...+ 1 a 4000 a 4001 =10  .
Question Image a 1 a 4001 = 4000d 10d =400  … (2)
From the equation (2), and we have
a 1 a 4001 =400 a 1 + a 4001 2 2 a 1 a 4001 2 2 =400  
By using (1), and we have
50 2 2 a 1 a 4001 2 2 =400 a 1 a 4001 2 2 = 25 2 400 a 1 a 4001 =2 225 =30  
Hence, the correct option is 2.
 
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