Question

If $a_1, a_2, a_3, \ldots . ., a_{4001}$ are terms of an A.P. such that $\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots \ldots+\frac{1}{a_{4000} a_{4001}}=10$ and $a_2+a_{4000}=50$, then find the value of $\left|a_1-a_{4001}\right|$.

Answer 1

Let common difference of A.P. $= d$
|Online test-4, P-2, 2
Now $\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots \ldots+\frac{1}{a_{4000} a_{4001}}=10$
$\Rightarrow \frac{1}{d}\left[\frac{\left(a_2-a_1\right)}{a_1 a_2}+\frac{a_3-a_2}{a_2 a_3}+\frac{a_4-a_3}{a_3 a_4}+\ldots .+\frac{a_{4001}-a_{4000}}{a_{4000} a_{4001}}\right]=10 \Rightarrow \frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{4001}}\right)=10$
$\Rightarrow \frac{a_{4001}-a_1}{d a_1 a_{4001}}=10 \Rightarrow \frac{4000 d}{d a_1 a_{4001}}=10 \Rightarrow a_1 a_{4001}=400$
But $a _2+ a _{4000}= a _1+ a _{4001}=50$
(As sum of terms equidistant from beginning and end in finiteA.P. is same)
Hence $\left(a_1-a_{4001}\right)^2=\left(a_1+a_{4001}\right)^2-4 a_1 a_{4001}=(50)^2-4 \times(400)=900 \Rightarrow\left|a_1-a_{4001}\right|=30$