Episode 8 — Aptitude and Reasoning / 8.16 — Arithmetic Progression

8.16.b Arithmetic Progression - Tips, Tricks and Shortcuts

Trick 1: Use Symmetric Selection to Avoid Variables

When a problem says "sum of 3 numbers in AP is X and their product is Y," always choose:

Three terms: (a - d), a, (a + d)

Sum = (a - d) + a + (a + d) = 3a
So: a = X / 3  (one variable eliminated immediately)

Then use the product condition to find d.

For 4 terms: Use (a - 3d), (a - d), (a + d), (a + 3d) with common difference 2d. For 5 terms: Use (a - 2d), (a - d), a, (a + d), (a + 2d).

Trick 2: Sum = Average x Number of Terms

For any AP:

Sum of AP = (Average of first and last term) x (Number of terms)
S(n) = [(a + l) / 2] * n

This is often faster than using the full formula. If you know the first and last term and the count, you are done.

Example: Sum of 3, 7, 11, 15, ..., 99

a = 3, l = 99, d = 4
n = (99 - 3)/4 + 1 = 25

Sum = (3 + 99)/2 * 25 = 51 * 25 = 1275

Trick 3: Quick nth Term Calculation

Instead of plugging into the formula, think of it as:

nth term = first term + (n - 1) jumps of d

"How many gaps from the first term to the nth term? It is (n-1) gaps."

Example: 7th term of 5, 9, 13, ...

6 gaps of 4 from the first term = 5 + 24 = 29

Trick 4: Finding Number of Terms Quickly

n = (last - first) / d + 1

Shortcut for multiples: How many multiples of 7 between 100 and 500?

First multiple of 7 >= 100: 105 (since 100/7 = 14.28... -> 15 * 7 = 105)
Last multiple of 7 <= 500: 497 (since 500/7 = 71.42... -> 71 * 7 = 497)

n = (497 - 105)/7 + 1 = 392/7 + 1 = 56 + 1 = 57

Trick 5: Sum of n Natural Numbers Shortcut

Memorize these results:

SumFormulaQuick Values
1 + 2 + ... + 1010*11/255
1 + 2 + ... + 2020*21/2210
1 + 2 + ... + 5050*51/21275
1 + 2 + ... + 100100*101/25050

Trick 6: Sum of Specific Range

Sum from the pth term to the qth term:

Sum from p to q = S(q) - S(p-1)

Alternative: Treat the sub-sequence as its own AP:

Number of terms = q - p + 1
First term of sub-sequence = a(p) = a + (p-1)d
Last term = a(q) = a + (q-1)d

Sum = (q - p + 1)/2 * [a(p) + a(q)]

Trick 7: The "Average = Middle Term" Rule

For an AP with an ODD number of terms:

Average of all terms = Middle term
Sum = Middle term * Number of terms

Example: Sum of 3, 7, 11, 15, 19 (5 terms)

Middle term = 11 (3rd term)
Sum = 11 * 5 = 55

For an AP with an EVEN number of terms:

Average = Average of the two middle terms
Sum = Average of two middle terms * Number of terms

Trick 8: Quick Checks for AP

Three numbers a, b, c are in AP if:

2b = a + c    (middle term = average of extremes)

This is faster than computing differences.

Example: Are 14, 23, 32 in AP?

2 * 23 = 46
14 + 32 = 46
Yes, they are in AP.

Trick 9: Sum of AP from S(n) Expression

If S(n) is given as a quadratic in n (like S(n) = An^2 + Bn), then:

d = 2A    (coefficient of n^2 times 2)
a = A + B  (sum of coefficients)

Example: S(n) = 3n^2 + 5n

d = 2 * 3 = 6
a = 3 + 5 = 8
AP: 8, 14, 20, 26, ...

Important: If S(n) has a constant term (like S(n) = 3n^2 + 5n + 2), the sequence is NOT an AP (since S(0) should be 0 for a proper AP).

Trick 10: Last Term When Sum is Given

If the sum of an AP equals a certain value S, and you know a and d, solve for n:

S = n/2 * [2a + (n-1)d]

This gives a quadratic in n:
d*n^2 + (2a - d)*n - 2S = 0

Use the quadratic formula and take the positive integer root.

Trick 11: Arithmetic Mean Shortcut

To insert k arithmetic means between a and b:

d = (b - a) / (k + 1)

The sum of all k means:

Sum = k * (a + b) / 2

This is because the k means plus the two endpoints form an AP, and the sum of the means equals k times the average of the extremes.

Trick 12: Common AP Sums to Memorize

SeriesSum
1 + 2 + 3 + ... + nn(n+1)/2
2 + 4 + 6 + ... + 2nn(n+1)
1 + 3 + 5 + ... + (2n-1)n^2
Sum of first n terms where a=1, d=1n(n+1)/2
Sum of AP: a, a+d, ... (n terms)n*a + n(n-1)d/2

Trick 13: Equidistant Terms Property

For quick problem solving, remember:

a(1) + a(n) = a(2) + a(n-1) = a(3) + a(n-2) = constant = a + l

Application: If the sum of the 3rd and 7th terms is 40, then the sum of the 1st and 9th terms, or the 4th and 6th terms, or the 5th term doubled, all equal 40.

Trick 14: Finding Common Difference When Two Terms Are Known

If the pth term is x and the qth term is y:

d = (y - x) / (q - p)
a = x - (p - 1) * d

Example: 5th term = 17, 10th term = 32

d = (32 - 17) / (10 - 5) = 15/5 = 3
a = 17 - (5-1)*3 = 17 - 12 = 5

Trick 15: Divisibility-Based Counting

How many numbers between A and B are divisible by k?

Count = floor(B/k) - floor((A-1)/k)

OR equivalently:
First term = ceil(A/k) * k
Last term = floor(B/k) * k
Count = (Last - First)/k + 1

Example: How many multiples of 6 between 15 and 100?

First = 18, Last = 96
Count = (96 - 18)/6 + 1 = 78/6 + 1 = 13 + 1 = 14

Exam Strategy Tips

Time Management

  1. Identify AP problems quickly: Look for keywords like "constant difference," "increases by same amount," "arithmetic sequence," "equally spaced."
  2. Formula first: Write down the formula before computing. Many errors come from wrong formula selection.
  3. Check your n: The most common mistake is miscounting the number of terms.

Common Traps

  1. Off-by-one error: n = (l-a)/d + 1, not (l-a)/d.
  2. Negative d: Don't forget that d can be negative. The formula still works.
  3. S(n) is not a(n): Students confuse the sum formula with the nth term formula.
  4. Fractional n: If n comes out as a fraction, either the value is not a term of the AP or there is a calculation error.

Verification Shortcuts

  • Check: a(1) = a (the first term from your formula should match)
  • Check: S(1) = a(1) (sum of 1 term = first term)
  • Check: S(2) - S(1) = a(2) (second term verification)

Next: 8.16.c - Solved Examples