besidedegree@gmail.com
+9779709005491
Back to Home
School SEE Compulsory Mathematics

Sequence and Series

Highlight Save
Arithmetic sequence: constant difference d; nth term = a + (n–1)d; sum Sₙ = n/2 (a + l) or n/2 [2a + (n–1)d]. Geometric sequence: constant ratio r; nth term = a r^(n–1); sum Sₙ = a(rⁿ–1)/(r–1) if r>1, or a(1–rⁿ)/(1–r) if r<1. Arithmetic means: one mean = (a+b)/2; k means → d = (b–a)/(k+1). Geometric means: one mean = √(ab); k means → r = (b/a)^(1/(k+1)).

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 term

7. 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

 

Visit this link for further practice!!

https://besidedegree.com/exam/s/academic

 

Related Videos

Arithmetic Progression (AP) Class 10 in Nepali || Algebra || Concept and Formula || By: Gurubaa

Important Links