Let the digits $a, b, c$ be in $A.P.$ Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in $A.P.$ at least once. How many such numbers can be formed?

The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is

If the $10^{\text {th }}$ term of an A.P. is $\frac{1}{20}$ and its $20^{\text {th }}$ term is $\frac{1}{10},$ then the sum of its first $200$ terms is

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 $TRUE$ ?

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

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

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

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

If ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $

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