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:

Recall the formula for the $n^{th}$ term of a sequence: \[ T_n = S_n - S_{n-1} \] where $S_n$ is the sum of the first $n$ terms. You can calculate $T_n$ by subtracting the sum of the first $(n-1)$ terms from the sum of the first $n$ terms.

Answer 1

The Correct Option is A

Step 1: Finding the general expression for $T_n$

The $n^{th}$ term $T_n$ of a progression can be found by subtracting the sum of the first $n-1$ terms from the sum of the first $n$ terms:

\[ T_n = S_n - S_{n-1} \]

Substituting the expression for $S_n$, we have:

\[ S_n = 2n^2 + 5n + 6 \]

Therefore, $S_{n-1}$ is:

\[ S_{n-1} = 2(n-1)^2 + 5(n-1) + 6 \]

Expanding $S_{n-1}$:

\[ S_{n-1} = 2(n^2 - 2n + 1) + 5(n - 1) + 6 \] \[ S_{n-1} = 2n^2 - 4n + 2 + 5n - 5 + 6 \] \[ S_{n-1} = 2n^2 + n + 3 \]

Now, we can find $T_n$ by subtracting $S_{n-1}$ from $S_n$:

\[ T_n = (2n^2 + 5n + 6) - (2n^2 + n + 3) \] \[ T_n = 2n^2 + 5n + 6 - 2n^2 - n - 3 \] \[ T_n = 4n + 3 \]

Thus, the $n^{th}$ term of the progression is:

\[ T_n = 4n + 3 \]

Step 2: Finding $\sum_{k=1}^{10} T_{3k}$

We need to find the sum of $T_{3k}$ for $k = 1$ to $10$. First, substitute $n = 3k$ into the expression for $T_n$:

\[ T_{3k} = 4(3k) + 3 = 12k + 3 \]

Now, we sum $T_{3k}$ for $k = 1$ to $10$:

\[ \sum_{k=1}^{10} T_{3k} = \sum_{k=1}^{10} (12k + 3) \]

We can break this into two separate sums:

\[ \sum_{k=1}^{10} T_{3k} = \sum_{k=1}^{10} 12k + \sum_{k=1}^{10} 3 \]

The first sum is the sum of the first 10 terms of $12k$:

\[ \sum_{k=1}^{10} 12k = 12 \sum_{k=1}^{10} k = 12 \times \frac{10(10+1)}{2} = 12 \times 55 = 660 \]

The second sum is simply:

\[ \sum_{k=1}^{10} 3 = 3 \times 10 = 30 \]

Now, add the two sums:

\[ \sum_{k=1}^{10} T_{3k} = 660 + 30 = 690 \]

Final Answer:

The sum $\sum_{k=1}^{10} T_{3k} = 690$