A collection of building blocks is set up into 51 rows. The first row has 124 blocks, and each row after that one has 4 more blocks than the preceding row. How many blocks are in the 10th row?

Answer 1

This problem describes an arithmetic sequence. Each row has a "term" representing the number of blocks in that row:

The equation for an arithmetic sequence is just like the equation for a line, y = mx + b. The "m" for the sequence is the growth rate (in this case 4 blocks each time equates to m = 4). The "b" (or y-intercept) is the number of blocks in the "zeroth row". But that row doesn't really exist - you just have to figure it out by going backwards from the 1st row. In this case, the "zeroth row" would have 120 blocks. So the equation is y = 4x + 120. We can adjust this equation to fit a sequence by changing the variables. Instead of "x", we can use "n" to represent the number of the row (or term number of the sequence). Instead of "y", we can use t(n) the represent the number of blocks (or term value in the sequence). So:

t(n) = 4n + 120

Try it for terms 1, 2, and 3:

t(1) = 4(1) + 120 = 124 (the 1st term is 124)

t(2) = 4(2) + 120 = 128 (the 2nd term is 128)

t(3) = 4(3) _ 120 = 132 (the 3rd term is 132)

So, to find the 10th term, we can plug in 10:

t(10) = 4(10) + 120 = 40 + 120 = 160 so the 10th term is 160 (or the 10th row has 160 blocks)

Answer 2

The first row has 124 blocks.

Row 2 has 124+4 = 128 blocks.

Row 3 has 128+4 = 132 blocks.

Row 10 will have 124 + 9*4 = 124 + 36 = 160 blocks.

The general rule is that the nth row has 124 + (n-1)*4 blocks.