Question

If for a progression, sum of the first $n$ terms is $2n^2+5n+6$ and $T_n$ is the $n^{th}$ term of the progression, then $\sum_{k=1}^{10} T_{3k} = $?

• A. $690$
• B. $270$
• C. $780$
• D. $460$

Hint:

• $T_1 = S_1$. For $n \ge 2$, $T_n = S_n - S_{n-1}$.
• Given $S_n = 2n^2+5n+6$.
• $T_1 = S_1 = 2(1)^2+5(1)+6 = 13$.
• For $n \ge 2$, $T_n = (2n^2+5n+6) - [2(n-1)^2+5(n-1)+6] = 4n+3$.
• We need $\sum_{k=1}^{10} T_{3k}$. The indices are $3, 6, \dots, 30$, all of which are $\ge 2$.
• So, $T_{3k} = 4(3k)+3 = 12k+3$.
• The sum is $\sum_{k=1}^{10} (12k+3) = 12 \sum_{k=1}^{10} k + \sum_{k=1}^{10} 3$.
• Use $\sum_{k=1}^{N} k = \frac{N(N+1)}{2}$. Sum = $12 \left(\frac{10 \cdot 11}{2}\right) + 3 \cdot 10 = 660 + 30 = 690$.

Answer 1

The Correct Option is A

Let $S_n$ be the sum of the first $n$ terms.
Given $S_n = 2n^2+5n+6$.
The $n^{th}$ term $T_n$ is $S_n - S_{n-1}$ for $n \ge 2$.
$T_n = (2n^2+5n+6) - (2(n-1)^2+5(n-1)+6)$ $T_n = (2n^2+5n+6) - (2(n^2-2n+1)+5n-5+6)$ $T_n = (2n^2+5n+6) - (2n^2-4n+2+5n-5+6)$ $T_n = (2n^2+5n+6) - (2n^2+n+3)$ $T_n = 4n+3$ for $n \ge 2$.
For $n=1$, $T_1 = S_1 = 2(1)^2+5(1)+6 = 2+5+6 = 13$.
(The formula $4n+3$ gives $T_1=7$, which is different from $S_1$.
This is expected when $S_n$ has a non-zero constant term and $S_0 \neq 0$ is not assumed).
We need to calculate $\sum_{k=1}^{10} T_{3k}$.
The terms are $T_3, T_6, T_9, \dots, T_{30}$.
All indices $3k$ are $\ge 2$.
So, for these terms, $T_{3k} = 4(3k)+3 = 12k+3$.
$\sum_{k=1}^{10} (12k+3) = \sum_{k=1}^{10} 12k + \sum_{k=1}^{10} 3$ $= 12 \sum_{k=1}^{10} k + 3 \sum_{k=1}^{10} 1$ $= 12 \left(\frac{10(10+1)}{2}\right) + 3(10)$ $= 12 \left(\frac{10 \cdot 11}{2}\right) + 30$ $= 12 (55) + 30 = 660 + 30 = 690$.
\[ \boxed{690} \]