1. Types of Sequence
TypeHow to IdentifyCommon Difference / Ration-th TermAPConstant differenced = t2 − t1tn = a + (n−1)dGPConstant ratior = t2 / t1tn = a × r^(n−1)2. Arithmetic Means (AP)
One mean between a and b:
AM = (a + b) / 2
k means between a and b:
Total terms = k + 2
Common difference: d = (b − a) / (k + 1)
Terms: a + d, a + 2d, …, a + k×d
Trick: Last term b = a + (k + 1)×d
3. Sum of AP (Arithmetic Series)
Steps:
Identify n, a, and l (first and last terms)
Use formula:
Sn = n/2 × (a + l)
or
Sn = n/2 × [2a + (n−1)d]
4. Geometric Means (GP)
One mean between a and b:
GM = √(a × b)
k means between a and b:
Total terms = k + 2
Solve r^(k+1) = b / a → r = (b / a)^(1/(k+1))
Terms: a × r, a × r^2, …, a × r^k
5. Sum of GP (Geometric Series)
Steps:
Identify a, r, n
Use formula:
If r > 1 → Sn = a × (r^n − 1)/(r − 1)
If r < 1 → Sn = a × (1 − r^n)/(1 − r)
Alternative → Sn = (last term × r − first term)/(r − 1)
6. Quick Formula Table
TopicFormulaWhen to UseAP nth termtn = a + (n−1)dFind any termAP sumSn = n/2 × (a + l) or n/2 × [2a + (n−1)d]Most exam questionsGP nth termtn = a × r^(n−1)Find any termGP sumSn = a × (r^n − 1)/(r−1)r > 1GP sumSn = a × (1 − r^n)/(1−r)r < 1One arithmetic mean(a + b)/2Insert one termOne geometric mean√(a × b)Insert one term7. Quick Tricks / Steps
Missing AP term → use middle term = average
Insert k AM → d = (b − a) / (k + 1)
Insert k GM → r = (b / a)^(1/(k+1))
Sum of first n natural numbers → n(n + 1)/2
Sum of AP when 2 terms given → find a and d first
Sum of GP when 2 terms given → find r first
Number of terms → solve quadratic from sum formula
Salary/production increasing every year → AP
Amount doubling/tripling → GP
Borrowed money in installments → GP (each installment = previous × r)
8. Solved Examples (Stepwise)
Q1: Insert 3 numbers between 4 and 24 (AP)
k = 3 → total terms = 5
d = (24 − 4)/4 = 5
Sequence = 4, 9, 14, 19, 24 → Inserted numbers: 9, 14, 19
Q2: Sum of 20 terms, 3 + 7 + 11 + …
a = 3, d = 4, n = 20
Last term l = a + (n−1)d = 3 + 19×4 = 79
Sn = n/2 × (a + l) = 10 × 82 = 820
Q3: Insert 3 numbers between 5 and 625 (GP)
k = 3 → total terms = 5
r^4 = 625 / 5 = 125 → r = 5
Sequence = 5, 25, 125, 625 → Inserted numbers: 25, 125
Q4: Sum of GP: 2 + 6 + 18 + … + 486
r = 3, n = 6
Sn = a × (r^n − 1)/(r − 1) = 2 × (3^6 − 1)/2 = 728
Q5: 4th term = 13, 10th term = 31 → sum of first 15 terms (AP)
d = (31 − 13)/6 = 3
a = t4 − 3d = 13 − 9 = 4
t15 = a + 14d = 4 + 42 = 46
S15 = 15/2 × (4 + 46) = 375
Q6: Insert 4 numbers between 10 and 50 (AP)
k = 4 → total terms = 6
d = (50 − 10)/5 = 8
Sequence = 10, 18, 26, 34, 42, 50 → Inserted numbers: 18, 26, 34, 42
Q7: Sum of first 25 terms, 7 + 12 + 17 + …
a = 7, d = 5, n = 25
Last term l = a + (n−1)d = 7 + 24×5 = 127
Sn = n/2 × (a + l) = 25/2 × (7 + 127) = 25 × 67 = 1675
Q8: Insert 2 numbers between 3 and 192 (GP)
k = 2 → total terms = 4
r^3 = 192 / 3 = 64 → r = 4
Sequence = 3, 12, 48, 192 → Inserted numbers: 12, 48
Q9: Sum of GP: 1 + 3 + 9 + … + 729
a = 1, r = 3
Last term = 729 → n = ? Solve 1 × 3^(n−1) = 729 → n−1 = 6 → n = 7
Sn = a × (r^n − 1)/(r − 1) = 1 × (3^7 − 1)/2 = 1093
Q10: 5th term = 20, 12th term = 41 → sum of first 15 terms (AP)
d = (41 − 20)/7 = 3
a = t5 − 4d = 20 − 12 = 8
t15 = a + 14d = 8 + 42 = 50
S15 = 15/2 × (8 + 50) = 435
Q11: Insert 3 numbers between 81 and 2187 (GP)
k = 3 → total terms = 5
r^4 = 2187 / 81 = 27 → r = 3
Sequence = 81, 243, 729, 2187 → Inserted numbers: 243, 729
Q12: Sum of first 12 terms, 5 + 8 + 11 + …
a = 5, d = 3, n = 12
Last term l = a + (n−1)d = 5 + 11×3 = 38
Sn = n/2 × (a + l) = 12/2 × (5 + 38) = 6 × 43 = 258
Q13: Insert 4 numbers between 2 and 162 (GP)
k = 4 → total terms = 6
r^5 = 162 / 2 = 81 → r = 3/2 = 1.5
Sequence = 2, 3, 4.5, 6.75, 10.125, 15.1875 → Inserted numbers: 3, 4.5, 6.75, 10.125
Q14: Sum of GP: 2 + 6 + 18 + … + 1458
a = 2, r = 3
Last term = 1458 → n = ? Solve 2 × 3^(n−1) = 1458 → n−1 = 6 → n = 7
Sn = a × (r^n − 1)/(r − 1) = 2 × (3^7 − 1)/2 = 2186
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