1. Types of Sequence
Description:
A sequence is an ordered list of numbers. Sequences can be Arithmetic (AP) or Geometric (GP).
AP: Each term differs from the previous by a constant difference.
GP: Each term is obtained by multiplying the previous term by a constant ratio.
Formulas:
AP: tₙ = a + (n−1)d
GP: tₙ = a × r^(n−1)
2. Arithmetic Means (AP)
Description:
Arithmetic means are numbers inserted between two given numbers to form an AP.
Formulas:
One mean: AM = (a + b)/2
k means: Total terms = k + 2, Common difference: d = (b − a)/(k + 1)
Terms: a + d, a + 2d, …, a + k×d
3. Sum of AP (Arithmetic Series)
Description:
The sum of terms in an AP can be calculated using either first & last terms or first term & common difference.
Formulas:
Sn = n/2 × (a + l)
or
Sn = n/2 × [2a + (n−1)d]
4. Geometric Means (GP)
Description:
Geometric means are numbers inserted between two numbers to form a GP.
Formulas:
One mean: GM = √(a × b)
k means: Solve r^(k+1) = b / a → r = (b / a)^(1/(k+1))
Terms: a × r, a × r², …, a × r^k
5. Sum of GP (Geometric Series)
Description:
The sum of n terms in a GP depends on the first term, common ratio, and number of terms.
Formulas:
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 Tricks / Steps
Missing AP term → 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
AP → salary, production, loan installments (linear increase)
GP → doubling, tripling, borrowed money with interest
Number of terms → solve quadratic from sum formula
7. Solved Examples
Q1: Insert 3 numbers between 4 and 24 (AP)
Solution:
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 + …
Solution:
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)
Solution:
k = 3 → total terms = 5
r⁴ = 625 / 5 = 125 → r = 5
Sequence = 5, 25, 125, 625
Inserted numbers = 25, 125
Q4: Sum of GP: 2 + 6 + 18 + … + 486
Solution:
a = 2
r = 3
Last term = 486
Find number of terms (n):
tₙ = a × r^(n−1)
486 = 2 × 3^(n−1)
3^(n−1) = 486 ÷ 2
3^(n−1) = 243
n − 1 = 5
n = 6
Sn = a × (r^n − 1) / (r − 1)
Sn = 2 × (3^6 − 1) / (3 − 1)
Sn = 2 × 728 / 2
Sn = 728
Q5: 4th term = 13, 10th term = 31 → sum of first 15 terms (AP)
Solution:
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)
Solution:
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 + …
Solution:
a = 7, d = 5, n = 25
Last term: l = a + (n−1)d = 7 + 24×5 = 127
Sn = n/2 × (a + l) = 25/2 × 134 = 1675
Q8: Insert 2 numbers between 3 and 192 (GP)
Solution:
k = 2 → total terms = 4
r³ = 192 / 3 = 64 → r = 4
Sequence = 3, 12, 48, 192
Inserted numbers = 12, 48
Q9: Sum of GP: 1 + 3 + 9 + … + 729
Solution:
a = 1
r = 3
Last term = 729
Find number of terms (n):
tₙ = a × r^(n−1)
729 = 1 × 3^(n−1)
3^(n−1) = 729
n − 1 = 6
n = 7
Sn = a × (r^n − 1) / (r − 1)
Sn = 1 × (3^7 − 1) / (3 − 1)
Sn = 1 × 2186 / 2
Sn = 1093
Q10: 5th term = 20, 12th term = 41 → sum of first 15 terms (AP)
Solution:
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)
Solution:
k = 3 → total terms = 5
r⁴ = 2187 / 81 = 27 → r = 3
Sequence = 81, 243, 729, 2187
Inserted numbers = 243, 729
Q12: Sum of first 12 terms: 5 + 8 + 11 + …
Solution:
a = 5, d = 3, n = 12
Last term: l = a + (n−1)d = 5 + 11×3 = 38
Sn = n/2 × (a + l) = 6 × 43 = 258
Q13: Insert 4 numbers between 2 and 162 (GP)
Solution:
k = 4 → total terms = 6
r⁵ = 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
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