Problem
Find a far-off term of a sequence
Row A runs 4, 11, 18, 25, 32, 39, …. Row B runs 253, 249, 245, 241, 237, …. Each row keeps going by its own fixed rule. Find the 25th number of each and say which is bigger, and by how much.
Reasoning · Grade 4-1 Number Sequences
Your answer
Watch out The 25th card tempts you to add 25 times, but the first card is free — only 24 jumps.
How to solve
Strategy Look for a Pattern — Writing out 25 cards for each row is possible but slow and easy to slip on. Instead I find the constant jump in each row, then count how many jumps separate the 1st card from the 25th. Listing the first few cards beside their positions makes the number of jumps visible, and testing the resulting rule on a card that is already printed proves the rule before it is trusted 25 cards out.
1STEP 1
Find the jump in each sequence
A goes up by 7, B goes down by 4.
A: +7 each step B: -4 each step
Checking the gap between every pair of neighbouring cards — not just the first pair — is what makes it safe to call the jump constant.
4.OA.C.5Look For A Pattern2STEP 2
Count the jumps, not the cards
Reaching the 25th takes 24 jumps.
jumps to the 25th card = 25 - 1 = 24
This is the same fencepost idea as counting the gaps between 25 posts in a row — there are always one fewer gaps than posts.
4.OA.C.5Make A Systematic ListReaching the 25th card takes 24 jumps, not 25, because a row always has one fewer gaps than it has cards.
Why?
Each jump sits between exactly one pair of neighbouring cards, so jumps and the gaps between cards match up one for one.
🧱One-to-one correspondenceIf each thing here pairs with exactly one thing there, the two groups are the same size.
Why?
Because every jump is the same size, the 25th card is the first card plus that jump repeated 24 times.
🧱A constant step reaches any termWhen a list climbs by the same amount every time, any term is the starting value plus that step repeated, so you never have to write out the ones between.
3STEP 3
Test the counting rule on a card that is already printed
Test the count on the printed 39 and 237 to check it works.
4 + 5 × 7 = 4 + 35 = 39 253 - 4 × 4 = 253 - 16 = 237
Checking the rule on a small case whose answer is already printed is proof the rule is right, and it costs only one multiplication.
4.NBT.B.5Solve An Easier Related Problem4STEP 4
Find the 25th number of sequence A
A's 25th is 4 + 24 × 7 = 172.
4 + 24 × 7 = 4 + 168 = 172
Twenty-four equal jumps of 7 is just 24 groups of 7, so a single multiplication replaces adding 7 twenty-four times.
4.NBT.B.5Look For A Pattern5STEP 5
Find the 25th number of sequence B
B's 25th is 253 − 24 × 4 = 157.
253 - 24 × 4 = 253 - 96 = 157
A shrinking sequence works the same way as a growing one; only the direction of the 24 jumps changes, so multiply then subtract.
4.NBT.B.5Look For A Pattern6STEP 6
Compare the two 25th numbers
A is bigger by 15.
172 - 157 = 15
Once both 25th numbers are written as plain whole numbers, comparing them is ordinary subtraction.
4.NBT.B.4Make A Systematic ListAnswer
A, 15 bigger
172 − 157 = 15
The two rows start far apart — B starts 249 above A — but they close on each other by 7 + 4 = 11 every step, so after 24 steps they close by 24 x 11 = 264. Starting 249 apart with B ahead and closing 264 means A ends up 264 - 249 = 15 ahead, exactly the difference found. The sizes are sensible too: A grows steadily from 4 to 172, and B falls from 253 to 157 while staying positive, so neither answer is out of range.
Takeaway
To reach the 25th number, don't count cards — count jumps: 24 of them, then multiply once instead of adding twenty-four times.
- Find the jump in each sequence
- Count the jumps, not the cards
- Test the counting rule on a card that is already printed
- Find the 25th number of sequence A
- Find the 25th number of sequence B
- Compare the two 25th numbers