Question

Let $a_1,a_2,\cdots ,a_{4001}$ are in A.P.. If $\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\cdots +\frac{1}{a_{4000}{a}_{4001}}=10$ and $a_2+a_{4000}=50,$ then

• A. $a_1{a}_{4001}=400$
• B. $a_1{a}_{4001}=401$
• C. $|a_1-a_{4001}|=40$
• D. $|a_1-a_{4001}|=30$

Answer 1

Answer: (A), (D)

The correct options are
A $a_1{a}_{4001}=400$
D $|a_1-a_{4001}|=30$

Let the common difference of the A.P. be $d$, so
$\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+\cdots +\frac{1}{a_{4000}{a}_{4001}}=10\Rightarrow \frac{1}{d}(\frac{a_2-a_1}{a_1{a}_2}+\frac{a_3-a_2}{a_2{a}_3}+\cdots +\frac{a_{4001}-a_{4000}}{a_{4000}{a}_{4001}})=10\Rightarrow \frac{1}{d}(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+\cdots +\frac{1}{a_{4000}}-\frac{1}{a_{4001}})=10\Rightarrow \frac{1}{d}(\frac{1}{a_1}-\frac{1}{a_{4001}})=10\Rightarrow \frac{1}{d}\left(\frac{a_{4001}-a_1}{a_1{a}_{4001}}\right)=10\Rightarrow \frac{4000}{a_1{a}_{4001}}=10(\because a_{4001}=a_1+4000d)\Rightarrow a_1{a}_{4001}=400$

Given, $a_2+a_{4000}=50$
$\Rightarrow a_1+a_{4001}=50\Rightarrow (a_1-a_{4001}{)}^2=(a_1+a_{4001}{)}^2-4a_1{a}_{4001}\Rightarrow |a_1-a_{4001}|=30$