We are given an arithmetic sequence $\{a_n\}$ with $a_2 = 6$ and $a_6 = 18$. We need to find the sum of the 4th term to the 11th term.

1. Problem Description

We are given an arithmetic sequence $\{a_n\}$ with $a_2 = 6$ and $a_6 = 18$. We need to find the sum of the 4th term to the 11th term.

2. Solution Steps

Let $a_1$ be the first term and $d$ be the common difference of the arithmetic sequence.

We have $a_2 = a_1 + d = 6$ and $a_6 = a_1 + 5d = 18$.

Subtracting the first equation from the second equation, we have:

$(a_1 + 5d) - (a_1 + d) = 18 - 6$

$4d = 12$

$d = 3$

Now, substitute $d = 3$ into the equation $a_1 + d = 6$:

$a_1 + 3 = 6$

$a_1 = 3$

Now we have $a_1 = 3$ and $d = 3$.

We want to find the sum of the 4th term to the 11th term, which is $S = a_4 + a_5 + \dots + a_{11}$.

The number of terms in the sum is $11 - 4 + 1 = 8$.

The sum of an arithmetic series is given by:

$S_n = \frac{n}{2}(a_1 + a_n)$

In our case, $n = 8$, $a_4 = a_1 + 3d = 3 + 3(3) = 3 + 9 = 12$, and $a_{11} = a_1 + 10d = 3 + 10(3) = 3 + 30 = 33$.

$S = \frac{8}{2}(a_4 + a_{11}) = 4(12 + 33) = 4(45) = 180$.

Alternatively, we can use the formula for the sum of an arithmetic series:

$S = \sum_{i=4}^{11} a_i = \sum_{i=4}^{11} (a_1 + (i-1)d) = \sum_{i=4}^{11} (3 + (i-1)3) = \sum_{i=4}^{11} (3i)$

$S = 3 \sum_{i=4}^{11} i = 3(4 + 5 + 6 + 7 + 8 + 9 + 10 + 11) = 3(70) = 210$

Another method is to use the sum of the first $n$ terms:

$S_n = \frac{n}{2}(2a_1 + (n-1)d)$

$S_{11} = \frac{11}{2}(2(3) + (11-1)3) = \frac{11}{2}(6 + 30) = \frac{11}{2}(36) = 11(18) = 198$

$S_3 = \frac{3}{2}(2(3) + (3-1)3) = \frac{3}{2}(6 + 6) = \frac{3}{2}(12) = 3(6) = 18$

$S = S_{11} - S_3 = 198 - 18 = 180$

3. Final Answer

The sum of the 4th term to the 11th term is

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8

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