If $a,\;b,\;c$ are in $A.P.$, then $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = $

  • A

    $1$

  • B

    $2$

  • C

    $3$

  • D

    $4$

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 $TRUE$ ?

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

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

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

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

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If $x_1 , x_2 ,  ..... , x_n$ and $\frac{1}{{{h_1}}},\frac{1}{{{h^2}}},......\frac{1}{{{h_n}}}$ are two $A.P' s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$, then $x_5. h_{10}$ equals

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If $x=\sum \limits_{n=0}^{\infty} a^{n}, y=\sum\limits_{n=0}^{\infty} b^{n}, z=\sum\limits_{n=0}^{\infty} c^{n}$, where $a , b , c$ are in $A.P.$ and $|a| < 1,|b| < 1,|c| < 1$, $abc \neq 0$, then

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The sum of all two digit numbers which, when divided by $4$, yield unity as a remainder is