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

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

Question
Let a1,a2,…,a4001a_1, a_2, \ldots, a_{4001} be in A.P. such that 1a1a2+1a2a3+⋯+1a4000a4001=10\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10 and a2+a4000=50a_2+a_{4000}=50. Then ∣a1−a4001∣|a_1-a_{4001}| equals
A2020
B3030correct
C4040
Dnone of these
Solution
Step 1: Let the common difference be dd. Since ak+1−ak=da_{k+1}-a_k=d,
1akak+1=1d(1ak−1ak+1).\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=1k=1 to 40004000; it telescopes:
∑k=140001akak+1=1d(1a1−1a4001)=1d⋅a4001−a1a1a4001.\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 a4001−a1=4000da_{4001}-a_1=4000d, so
4000a1a4001=10 ⇒ a1a4001=400.\frac{4000}{a_1 a_{4001}}=10\ \Rightarrow\ a_1 a_{4001}=400.
 Step 4: Terms equidistant from the ends have equal sum: a1+a4001=a2+a4000=50.a_1+a_{4001}=a_2+a_{4000}=50. Step 5:
∣a1−a4001∣2=(a1+a4001)2−4a1a4001=2500−1600=900 ⇒ ∣a1−a4001∣=30.|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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