From two towns 507km apart, Dave and Bill set out to meet each other. Dave travels 1km the first day, 3 the second, 5 the third, and so on in an arithmetic sequence, while Bill travels 2km the first, 6 the second, 10 the third, etc. How many days after they started will they meet?

Answer 1

you want the total distance traveled by both to be at least 507. So, using your two sequences,

n/2 (2*1 + (n-1)*2) + n/2 (2*2 + (n-1)*2) >= 507

Answer 2

You have two arithmetic sequences.

Dave´s sequence 1 , 3 , 5 ...

the initial term of a1 = 1 and common difference d = 2

nth term of the sequence:

an = a1 + ( n - 1 ) ∙ d

an = 1 + ( n - 1 ) ∙ 2

an = 1 + 2 n - 2

an = 2 n - 1

The sum of the n members of a arithmetic sequence:

Sn = ( n / 2 ) ( a1 + an )

Sn = ( n / 2 ) ( 1 + 2 n - 1 )

Sn = ( n / 2 ) ∙ 2 n

Sn = n²
for Dave

Bill´s sequence 2 , 6 , 10 ...

the initial term of a1 = 2 and common difference d = 4

nth term of the sequence

an = a1 + ( n - 1 ) ∙ d

an = 2 + ( n - 1 ) ∙ 4

an = 2 + 4 n - 4

an = 4 n - 2

The sum of the n members of a arithmetic sequence:

Sn = ( n / 2 ) ( a1 + an )

Sn = ( n / 2 ) ( 2 + 4 n - 2 )

Sn = ( n / 2 ) ∙ 4 n

Sn = 2 n²
for Bill

The combined distance = 507 km so:

n² + 2 n² = 507

3 n² = 507

n² = 507 / 3

n² = 169

n = ± √169

n = ± 13

Number of days can't be negative so:

n = 13

Prooof:

Dave travel:

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 169 km

Bill travel:

2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 38 + 42 + 46 + 50 = 338 km

169 km + 338 km = 507 km

Answer 3

My bad. I was using the sequence 2,4,6,8...
rather than 2,6,10,14,...
go with Bosnian.