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AP Reciprocal Sum: Value of |a1 - a4001| | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

a_1, a_2, \ldots, a_{4001}

be in A.P. such that

\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10

and

a_2+a_{4000}=50

. Then

|a_1-a_{4001}|

equals

20

30

correct

40

Dnone of these

Solution

Step 1: Let the common difference be

d

. Since

a_{k+1}-a_k=d

\frac{1}{a_k a_{k+1}}=\frac{1}{d}\left(\frac{1}{a_k}-\frac{1}{a_{k+1}}\right).

Step 2: Sum from

k=1

4000

; it telescopes:

\sum_{k=1}^{4000}\frac{1}{a_k a_{k+1}}=\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{4001}}\right)=\frac{1}{d}\cdot\frac{a_{4001}-a_1}{a_1 a_{4001}}.

Step 3: Here

a_{4001}-a_1=4000d

, so

\frac{4000}{a_1 a_{4001}}=10\ \Rightarrow\ a_1 a_{4001}=400.

Step 4: Terms equidistant from the ends have equal sum:

a_1+a_{4001}=a_2+a_{4000}=50.

Step 5:

|a_1-a_{4001}|^2=(a_1+a_{4001})^2-4a_1 a_{4001}=2500-1600=900\ \Rightarrow\ |a_1-a_{4001}|=30.

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