In the sequence 1, 7, 13, 19, . . . , each number is 6 more than the number before it. In the sequence 1, 9, 17, 25, . . . , each number is 8 more than the number before it. The two sequences have infinitely many numbers in common. Find the sum of the first three common numbers.

first three common numbers are

25, 73 and 97

And their sum= 195.

Step-by-step explanation:

1, 7, 13, 19, . . adding 6 to the terms in this first sequence gives the first 17 terms as

1,7,13,19,25,31,37,43,49,55,61,67,73,79,85,91,97 and

In the second sequence the first 17 terms with the addition of 8 are

1,9,17,25,33,41,49,57,65,73,81,89,97,105, 113,121,129,137

From the above, we can see common terms so far from the two sequence as

25, 73 and 97

Adding them gives a total of 195.

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

answer: the answer is simply nothing.

step-by-step explanation: because there is no ending to this question

gimme brainliest > : 3