Find three consecutive odd integers whose sum is 117

First odd integer: 2k+1 (note: using 2k+1 insures it is an odd integer)

Second odd integer: 2k+3

Third odd integer: 2k+5

First + Second + Third = 117

2k+1 + 2k+3 + 2k+5 = 117 substituted for First, Second, and Third

6k + 9 = 117 added like terms

6k = 108 subtracted 7 from both sides

k = 18

First: 2k + 1 = 2(18) + 1 = 36 + 1 = 37

Second: 2k + 3 = 2(18) + 3 = 36 + 3 = 39

Third: 2k + 5 = 2(18) + 5 = 36 + 5 = 41

37, 39, 41

the answer is b; false. 2 dimensional things are a lot more different from 3 dimensional things.

i hope this ! : d

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= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5