Answer:
The formula is
![a_n = 3n-2](https://img.qammunity.org/2022/formulas/mathematics/high-school/mvl6ouqi29tx9cd0jywc82np2a3dlx1wwo.png)
The 14th term is 40, in other words,
![a_(14) = 40](https://img.qammunity.org/2022/formulas/mathematics/high-school/e6a77cztns06y3i3jhji0dpl9a36q8bweg.png)
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Step-by-step explanation:
The given sequence is 1, 4, 7, 10, ...
Subtract each adjacent term:
This shows that each term is increasing by 3. This is the common difference, so d = 3.
The first term is
![a_1 = 1](https://img.qammunity.org/2022/formulas/mathematics/college/yn78hhreglomcvt1q0ki42gw48nlhctkrc.png)
The nth term of the arithmetic sequence is...
![a_n = a_1 + d(n-1)\\\\a_n = 1 + 3(n-1)\\\\a_n = 1 + 3n-3\\\\a_n = 3n-2\\\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/d8fy4oy5zxn9h6rpkukddr841ndx2i7k1m.png)
As a check, let's plug in say n = 3 to find that:
![a_n = 3n-2\\\\a_3 = 3(3)-2\\\\a_3 = 9-2\\\\a_3 = 7\\\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/nszxf1rhwcvv8sqxwfy6ulxn2jli70gyay.png)
The third term is 7, which matches with the sequence given to us. I'll let you confirm the other values.
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We'll repeat this idea but now for n = 14 to find the 14th term of the sequence.
![a_n = 3n-2\\\\a_(14) = 3(14)-2\\\\a_(14) = 42-2\\\\a_(14) = 40\\\\](https://img.qammunity.org/2022/formulas/mathematics/high-school/6eot0w3ygzjhd4w014erbvw95g3ntd3iyk.png)
The 14th term is 40.
You can make a table of values to help confirm the answer. In the first column you'll have the values of n (1,2,3,...) all the way up to 14. The second column will consist of the sequence 1, 4, 7, 10, ... each time going up by 3 until you reach the 14th term of 40.