Question

If $a_1,a_2,a_3,\dots ..,a_{4001}$ are terms of an A.P. such that $\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\dots \dots +\frac{1}{a_{4000}{a}_{4001}}=10$ and $a_2+a_{4000}=50$, then find the value of $\left|a_1-a_{4001}\right|$.

Answer 1

Answer: 30

Let common difference of A.P. $=d$
|Online test-4, P-2, 2
Now $\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\dots \dots +\frac{1}{a_{4000}{a}_{4001}}=10$
$\Rightarrow \frac{1}{d}\left[\frac{\left(a_2-a_1\right)}{a_1{a}_2}+\frac{a_3-a_2}{a_2{a}_3}+\frac{a_4-a_3}{a_3{a}_4}+\dots .+\frac{a_{4001}-a_{4000}}{a_{4000}{a}_{4001}}\right]=10\Rightarrow \frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{4001}}\right)=10$
$\Rightarrow \frac{a_{4001}-a_1}{d{a}_1{a}_{4001}}=10\Rightarrow \frac{4000d}{d{a}_1{a}_{4001}}=10\Rightarrow a_1{a}_{4001}=400$
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 ${\left(a_1-a_{4001}\right)}^2={\left(a_1+a_{4001}\right)}^2-4a_1{a}_{4001}=(50{)}^2-4\times (400)=900\Rightarrow \left|a_1-a_{4001}\right|=30$