3 + x, 9 + 3x, 13 + 4x, ... is an arithmetic sequence for some real number x. b. Find the 10th term of the sequence.

Answer: The 10th term is 19Step-by-step explanation:For a sequence to be an arithmetic sequence, then there exist a common difference. [tex]\\[/tex]How do we calculate common difference? [tex]\\[/tex]Common difference is calculated by subtracting the first term from the second term, subtracting the second term from the third term and so on. [tex]\\[/tex]From the sequence given , 3 + x, 9 + 3x, 13 + 4x, .. , calculating common difference implies [tex]\\[/tex](9+3x) – ( 3 +x) = 13 + 4x – ( 9 + 3x) [tex]\\[/tex]Expanding, we have [tex]\\[/tex]9 + 3x – 3 – x = 13 + 4x – 9 – 3x [tex]\\[/tex]⇒9 – 3+3x –x = 13 – 9 +4x – 3x [tex]\\[/tex]⇒6 + 2x = 4 + x [tex]\\[/tex]Collecting the like terms  , we have[tex]\\[/tex]2x – x = 4 – 6 [tex]\\[/tex] which gives X = -2 [tex]\\[/tex]Since the value of x is gotten now , we can get what the given sequence looks like by substituting the value of x into each term given, the sequence becomes: [tex]\\[/tex]3 + (-2), 9 + 3(-2), 13 + 4(-2) +...[tex]\\[/tex]Which gives [tex]\\[/tex]1 , 3 ,5, … [tex]\\[/tex]We can therefore note that the common difference is 2 [tex]\\[/tex]check: 3 - 1 = 5 -3 = 2[tex]\\[/tex](b) To calculate the tenth term , we use the general formula for calculating nth term , which is given as[tex]\\[/tex][tex]T_{n}[/tex] = a + (n-1)d [tex]\\[/tex]Where n is the number of terms , a is the first term and d is the common difference,  [tex]\\[/tex]Therefore, [tex]T_{10}[/tex] = 1 + ( 10-1)x2 [tex]\\[/tex]= 1 + 18 [tex]\\[/tex] = 19 Therefore the 10th term is 19