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Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
Which one of the following is a correct statement?
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Let $V_r$ denotes the sum of the first $r$ terms of an arithmetic progression $(A P)$ whose first term is $r$ and the common difference is $(2 r-1)$. Let $T_r=V_{r+1}-V_r-2$ and $Q_r=T_{r+1}-T_r$ for $r=1,2, \ldots$Question:
Which one of the following is a correct statement?
Solution:
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Verified Answer
The correct answer is:
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 6
$Q_1, Q_2, Q_3, \ldots$ are in AP with common difference 6
Since, $T_r=3 r^2+2 r-1$
$$
\begin{array}{rlrl}
& \therefore & T_{r+1} & =3(r+1)^2+2(r+1)-1 \\
& \therefore & Q_r & =T_{r+1}-T_r=3[2 r+1]+2[1] \\
\Rightarrow & Q_r & =6 r+5 \\
\Rightarrow & Q_{r+1} & =6(r+1)+5
\end{array}
$$
Common difference $=Q_{r+1}-Q_r=6$
$$
\begin{array}{rlrl}
& \therefore & T_{r+1} & =3(r+1)^2+2(r+1)-1 \\
& \therefore & Q_r & =T_{r+1}-T_r=3[2 r+1]+2[1] \\
\Rightarrow & Q_r & =6 r+5 \\
\Rightarrow & Q_{r+1} & =6(r+1)+5
\end{array}
$$
Common difference $=Q_{r+1}-Q_r=6$
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