Question16. Given that the three expressions below belong to the first three terms of an arithmetic sequence, find the following: a) Find the value of the of " $x$ " (1 Point) $t_{1}=3 x \quad t_{2}=4 x+2 \quad t_{3}=6 x-2$
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17. b) FInd the value of the common difference "d" (1 Point)
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18. c) Find the sum of the first 10 terms ( 1 Point)
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Studdy Solution

STEP 1

1. The expressions $t_1 = 3x$ , $t_2 = 4x + 2$ , and $t_3 = 6x - 2$ are the first three terms of an arithmetic sequence.
2. In an arithmetic sequence, the difference between consecutive terms is constant and is called the common difference $d$ .
3. The sum of the first $n$ terms of an arithmetic sequence can be calculated using the formula $S_n = \frac{n}{2} (2a + (n-1)d)$ , where $a$ is the first term and $d$ is the common difference.

STEP 2

1. Find the value of $x$ .
2. Find the value of the common difference $d$ .
3. Find the sum of the first 10 terms.

STEP 3

To find the value of $x$ , we use the property of arithmetic sequences that the difference between consecutive terms is constant. Therefore, the difference between $t_2$ and $t_1$ should be equal to the difference between $t_3$ and $t_2$ .
Calculate the difference $t_2 - t_1$ :
$t_2 - t_1 = (4x + 2) - 3x$ $t_2 - t_1 = x + 2$
Calculate the difference $t_3 - t_2$ :
$t_3 - t_2 = (6x - 2) - (4x + 2)$ $t_3 - t_2 = 2x - 4$
Set the two differences equal to each other:
$x + 2 = 2x - 4$
Solve for $x$ :
$2 + 4 = 2x - x$ $6 = x$

STEP 4

Now that we have $x = 6$ , substitute $x$ back into the expressions to find the terms:
$t_1 = 3x = 3(6) = 18$ $t_2 = 4x + 2 = 4(6) + 2 = 26$ $t_3 = 6x - 2 = 6(6) - 2 = 34$
Calculate the common difference $d$ :
$d = t_2 - t_1 = 26 - 18 = 8$

STEP 5

To find the sum of the first 10 terms, use the formula for the sum of an arithmetic sequence:
$S_n = \frac{n}{2} (2a + (n-1)d)$
Here, $n = 10$ , $a = t_1 = 18$ , and $d = 8$ .
$S_{10} = \frac{10}{2} (2 \times 18 + (10-1) \times 8)$ $S_{10} = 5 (36 + 72)$ $S_{10} = 5 \times 108$ $S_{10} = 540$
The value of $x$ is:
$\boxed{6}$
The value of the common difference $d$ is:
$\boxed{8}$
The sum of the first 10 terms is:
$\boxed{540}$

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