1. Arithmetic Progressions
An AP adds the same fixed amount (the common difference,
d) each step. Both formulas below can be re-derived on the spot if
forgotten — worth doing once so they're never a pure memorization risk.
Sequence: a, a+d, a+2d, a+3d, ...
The nth term has been "stepped" (n-1) times from the first term:
a_n = a + (n-1)d
Worked: 5, 8, 11, 14, ... find the 20th term.
a=5, d=3
a_20 = 5 + 19x3 = 5 + 57 = 62Sum of the first n terms, derived by pairing the first and
last term, second and second-last, etc:
S = a + (a+d) + (a+2d) + ... + a_n
S = a_n + (a_n-d) + (a_n-2d) + ... + a (written backward)
Adding both: 2S = n x (a + a_n)
S = n/2 x (a + a_n) -- "average of first and last, times count"
Worked: sum of 5, 8, 11, ..., 62 (20 terms, from above)
S = 20/2 x (5+62) = 10 x 67 = 670"Average of first and last term, times the number of terms" is worth holding onto as the intuition, not just the formula — it's literally what an average times a count means, applied to a sequence that happens to be evenly spaced.
2. Geometric Progressions
A GP multiplies by the same fixed ratio (r) each step — Week 5's
compound interest is a GP in disguise, with the interest rate as the common ratio.
Sequence: a, ar, ar^2, ar^3, ...
nth term: a_n = a x r^(n-1)
Sum of first n terms (for r != 1):
S = a(r^n - 1) / (r - 1) [if r > 1]
S = a(1 - r^n) / (1 - r) [if r < 1 -- same formula,
algebraically rearranged
to avoid a negative/negative]
Worked: 3, 6, 12, 24, ... find the sum of the first 8 terms.
a=3, r=2
S = 3(2^8 - 1)/(2-1) = 3 x 255 = 7653. The Infinite GP Sum
A GP with |r| < 1 has terms shrinking toward zero — summed forever,
the total still converges to a finite number, which is the single most useful
special case in this chapter.
S_infinity = a / (1 - r) [ONLY valid when -1 < r < 1]
Worked: find the sum of 1/2 + 1/4 + 1/8 + 1/16 + ... forever
a = 1/2, r = 1/2
S = (1/2) / (1 - 1/2) = (1/2)/(1/2) = 1
An infinite sum of ever-smaller pieces sums to exactly 1 --
makes intuitive sense: repeatedly halving a remaining gap
approaches, but never exceeds, the whole.
This formula is invalid — and meaningless — for |r| ≥ 1, since the
terms don't shrink and the sum genuinely grows without bound. A common recurring
decimal question ("convert 0.777... to a fraction") is exactly this formula in
disguise: 0.777... = 7/10 + 7/100 + 7/1000 + ... is an infinite GP with a=7/10,
r=1/10, summing to (7/10)/(1-1/10) = 7/9.
4. Spotting a Series' Pattern
Not every series is a clean AP or GP — test questions frequently mix operations, or use squares, alternating patterns, or a pattern applied to differences rather than the terms themselves.
1. Check CONSECUTIVE DIFFERENCES: constant -> AP
2, 5, 8, 11 -> differences: 3,3,3 -> AP, d=3
2. Check CONSECUTIVE RATIOS: constant -> GP
2, 6, 18, 54 -> ratios: 3,3,3 -> GP, r=3
3. If neither is constant, check the DIFFERENCES of the
DIFFERENCES (second-order pattern):
1, 2, 4, 7, 11 -> differences: 1,2,3,4 -> THOSE differ
by a constant 1 -> the original series is a "second-order"
pattern, not a simple AP -- but still solvable by extending
the difference pattern
4. Watch for alternating or paired patterns:
1, 4, 2, 8, 3, 12, 4, 16... -> odd positions: 1,2,3,4 (AP);
even positions: 4,8,12,16 (AP) -- TWO interleaved series,
not oneThe consistent approach: try Step 1 first, then Step 2, then Step 3, then check for interleaving — in roughly that order of likelihood on most tests. Spending ten seconds running through this checklist beats staring at a series hoping the pattern jumps out.
5. Practice Problems
30 problems, no calculator — Easy, Medium & Tough
Run Section 4's diagnostic checklist on any series before assuming its type. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.
Easy (1–10)
- Find the 5th term of the AP: 2, 5, 8, 11, ...
- Find the common difference of the AP: 10, 7, 4, 1, ...
- Find the sum of the first 5 terms of the AP: 1, 2, 3, 4, 5.
- Find the 4th term of the GP: 2, 6, 18, 54, ...
- Find the common ratio of the GP: 5, 10, 20, 40, ...
- Find the 10th term of the AP: 3, 7, 11, ...
- Find the sum of the first 10 natural numbers.
- Is the sequence 4, 8, 12, 16 an AP or a GP?
- Is the sequence 3, 9, 27, 81 an AP or a GP?
- Find the sum to infinity of 1 + 1/2 + 1/4 + 1/8 + ...
1) a=2,d=3; a5=2+4×3=14. 2) d=-3. 3) 15. 4) a=2,r=3; a4=2×27=54. 5) r=2. 6) a10=3+9×4=39. 7) 55. 8) AP (d=4). 9) GP (r=3). 10) S=1/(1-0.5)=2.
Medium (11–20)
- Find the sum of the first 20 terms of the AP: 2, 4, 6, 8, ...
- Find the 8th term of the AP whose first term is 5 and common difference is 3.
- Find the sum of the first 6 terms of the GP: 3, 6, 12, 24, ...
- The 5th term of an AP is 23 and the 9th term is 39. Find the common difference.
- Find the sum of all multiples of 5 between 1 and 100.
- Find the sum to infinity of 5 + 5/3 + 5/9 + ...
- Find the number of terms in the AP 7, 10, 13, ..., 100.
- The sum of the first n terms of an AP is 3n² + 2n. Find the first term.
- Find the 6th term of the GP whose first term is 4 and common ratio is 3.
- Convert the recurring decimal 0.4545... to a fraction using the infinite GP formula.
11) 2×(1+2+...+20)=2×210=420. 12) a8=5+7×3=26. 13) a=3,r=2,n=6; S=3(63)=189. 14) 4d=16 → d=4. 15) 20 terms; S=20/2×(5+100)=1,050. 16) a=5,r=1/3; S=5/(2/3)=7.5. 17) (100-7)/3+1=32 terms. 18) S(1)=3+2=5. 19) a6=4×3⁵=972. 20) 45/99=5/11.
Tough (21–30)
- Find the sum of the first 15 terms of the AP: 5, 8, 11, ...
- The sum of the first n terms of an AP is 4n² - n. Find the 10th term.
- Find the next two terms of the series: 1, 4, 9, 16, 25, ...
- Find the next two terms of the series: 2, 6, 12, 20, 30, ...
- The 3rd and 6th terms of a GP are 12 and 96 respectively. Find the first term and common ratio.
- Find the sum of the first 10 terms of the series 1, 3, 5, 7, ... (first 10 odd numbers).
- If the sum of three numbers in AP is 24 and their product is 440, find the numbers.
- Find the sum to infinity of the GP: 8, -4, 2, -1, ...
- The sum of the first n terms of an AP is 5n² + 3n. Find the common difference of the AP.
- Three numbers are in GP such that their sum is 26 and their product is 216. Find the numbers.
21) a=5,d=3,a15=47; S=15/2×(5+47)=390. 22) S(10)=390, S(9)=315 → a10=75. 23) Differences 3,5,7,9 → next diffs 11,13 → terms 36, 49. 24) Differences 4,6,8,10 → next diffs 12,14 → terms 42, 56. 25) r³=96/12=8 → r=2; a=12/4=3. 26) Sum of first n odd numbers=n²=100. 27) a-d,a,a+d; sum=3a=24 → a=8; product 8(64-d²)=440 → d²=9 → d=3; numbers 5,8,11. 28) a=8,r=-1/2; S=8/1.5=16/3≈5.33. 29) a1=S(1)=8; a2=S(2)-S(1)=26-8=18; d=10. 30) a³=216 → a=6; 6/r+6+6r=26 → 3r²-10r+3=0 → r=3 or 1/3; numbers 2,6,18.
6. Knowledge Check
Four quick questions. Expand each to check your answer.
Q1
Why does the AP sum formula S = n/2 × (a + last term) work — what's the intuition behind pairing first-with-last?
Why does the AP sum formula S = n/2 × (a + last term) work — what's the intuition behind pairing first-with-last?
Writing the sum forward and backward and adding the two versions term-by-term makes every pair (first+last, second+second-last, etc.) sum to the exact same total, because whatever one term loses relative to the middle, its paired partner gains. This produces n copies of that constant pair-sum, giving 2S = n×(a+last), which simplifies to the average-times-count formula.
Q2
Why is the infinite GP sum formula only valid for |r| < 1, not for r ≥ 1?
Why is the infinite GP sum formula only valid for |r| < 1, not for r ≥ 1?
Only when |r| < 1 do the terms shrink toward zero as the sequence continues, allowing the running total to approach a finite limiting value rather than growing forever. For r ≥ 1, each term is the same size or larger than the last, so the sum genuinely diverges to infinity, and no finite value can correctly describe it.
Q3
For the series 1, 2, 4, 7, 11, why does checking the differences of the differences help identify the pattern, when the first-level differences aren't constant?
For the series 1, 2, 4, 7, 11, why does checking the differences of the differences help identify the pattern, when the first-level differences aren't constant?
The first-level differences themselves (1, 2, 3, 4) form their own arithmetic progression, meaning the original series isn't a simple AP but has a consistent underlying structure one level deeper. Recognizing that the differences form a recognizable pattern (here, a simple AP) lets you extend the original series correctly by first extending the difference pattern.
Q4
Why does treating 1, 4, 2, 8, 3, 12, 4, 16 as one single AP or GP fail, even though it clearly follows some pattern?
Why does treating 1, 4, 2, 8, 3, 12, 4, 16 as one single AP or GP fail, even though it clearly follows some pattern?
The series is actually two separate, interleaved sequences (odd positions: 1,2,3,4; even positions: 4,8,12,16), each independently a clean AP, but merged together they don't have a single constant difference or ratio when read straight through. Recognizing the interleaving and separating it into its two component series is necessary before either can be analyzed as an AP or GP.