+0

# How many terms in a sequence of numbers. Thanks for helping!

0
122
2

How many numbers are in the list 6, 7, 10, 11, 14, 15,...,94, 95, 98?

There are incrememnts of three every set of two numbers, but I'm not sure how to continue.

May 17, 2020

#1
0

listfor(n, 1,24, (4*n+2, 4*n+3))=(6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99)

May 17, 2020
#2
+21959
0

You can use this formula to help you find the number of terms:  tn  =  t1 + (n - 1)d

tn  is the last term of an arithmetic sequence.

t1  is the first term of an arithemtic sequence

n   is the number of terms

d   is the common difference

This sequence contains two arithmetic sequences:

6, 10, 14, 18, ... 94, 98          and        7, 11, 15, ... 91, 95

In both of these sequences, d = 4.

For  6, 10, 14, 18, ... 94, 98                                   tn  =  t1 + (n - 1)d

tn  =  98                                                                 98  =  6 + (n - 1)4

t1  = 6                                                                    92  =  4n - 4

n   =  unknown                                                       96  =  4n

d    =  4                                                                   24  =  n

For  7, 11, 15, ... 91, 95                                         tn  =  t1 + (n - 1)d

tn  =  95                                                                 95  =  7+ (n - 1)4

t1  = 7                                                                    88  =  4n - 4

n   =  unknown                                                       92  =  4n

d    =  4                                                                   23  =  n

So, there will be a total of  24 + 23  =  47 terms

May 17, 2020