Question

Let $\left\{a_n\right\}$ and $\left\{T_n\right\}$ be two different sequences such that $a_{n+1}=2+a_n\forall n\in N,a_{19}=21$ and $T_n=\frac{1}{a_n{a}_{n+2}}\forall n\in N,$ then value of $\sum _{r=1}^{\infty }{T}_r$

• A. $\frac{2}{15}$
• B. $\frac{7}{195}$
• C. $\frac{1}{6}$
• D. $\frac{8}{15}$

Answer 1

The correct option is B $\frac{7}{195}$

$given{a}_{n+1}=a_n+2$

$Let{a}_1=x,a_2=2+x,a_3=4+x$

$a_n=2(n-1)+x$

$a_{19}=21$

$21=36+x$

$x=-15$

$a_1=-15,a2=2+a_1=-13$

$Let{\sum }_1^{\infty }{T}_r=S$

$\frac{1}{a_n{a}_{n+2}}=\frac{1}{a_{n+2}-a_n}[\frac{1}{a_n}-\frac{1}{a_{n+2}}]$

$a_{n+2}-a_n=4$

$S=\left(1/4\right)\ast [\frac{1}{a_1}-\frac{1}{a_3}+\frac{1}{a_2}-\frac{1}{a_4}+\frac{1}{a_3}-\frac{1}{a_5}+\frac{1}{a_4}-\frac{1}{a_6}+\frac{1}{a_5}-.......]$

$Fromtheaboveequation,S=\left(1/4\right)\ast (1/a_1+1/a_2)=\left(1/4\right)\ast (-1/15-1/13)=7/195$