An arithmetic progression is written in the following way. The sum of all the terms of the $10^{\text{th}}$ row is..........

  • A
    $1505$
  • B
    $1078$
  • C
    $1045$
  • D
    $1548$

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Let $a_{1}, a_{2}, \ldots, a_{10}$ be an $AP$ with common difference $-3$ and $b_{1}, b_{2}, \ldots, b_{10}$ be a $GP$ with common ratio $2$. Let $c_{k}=a_{k}+b_{k}, k=1, 2, \ldots, 10$. If $c_{2}=12$ and $c_{3}=13$,then $\sum_{k=1}^{10} c_{k}$ is equal to:

Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

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