Let ${S_1},{S_2},......,{S_{101}}$ be the consecutive terms of an $A.P$ . If $\frac{1}{{{S_1}{S_2}}} + \frac{1}{{{S_2}{S_3}}} + .... + \frac{1}{{{S_{100}}{S_{101}}}} = \frac{1}{6}$ and ${S_1} + {S_{101}} = 50$ , then $\left| {{S_1} - {S_{101}}} \right|$ is equal to

  • A

    $10$

  • B

    $20$

  • C

    $30$

  • D

    $40$

Similar Questions

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$ . Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then, which of the following is/are $TRUE$ ?

$(A)$ $T_{20}=1604$

$(B)$ $\sum_{ k =1}^{20} T_{ k }=10510$

$(C)$ $T_{30}=3454$

$(D)$ $\sum_{ k =1}^{30} T_{ k }=35610$

  • [IIT 2022]

If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in $A.P.,$ prove that $a, b, c$ are in $A.P.$

The four arithmetic means between $3$ and $23$ are

The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6 .$ Find the ratio of their $18^{\text {th }}$ terms.

The Fibonacci sequence is defined by

$1 = {a_1} = {a_2}{\rm{ }}$ and ${a_n} = {a_{n - 1}} + {a_{n - 2}},n\, > \,2$

Find $\frac{a_{n+1}}{a_{n}},$ for $n=1,2,3,4,5$