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#NMMS37 | SERIES | Arithmetic Progression | Sum of AP | National Means cum Merits Scholarship Exam |


In this video, we solve a Mathematics Problem from Arithmetic Progression (A.P.) / Geometric Progression (G.P.) involving Surds (Square Roots).
The given sequence is:
√2, √8, √18, √32, …
We are asked to find the sum of the first 10 terms of this series.

At first glance, this might seem like a complex square root series — but the key is to simplify each term correctly and identify the pattern.

Step-by-Step Solution:

Let’s simplify the given terms:
√2, √8, √18, √32, …

Now,
√8 = √(4×2) = 2√2
√18 = √(9×2) = 3√2
√32 = √(16×2) = 4√2

Hence, the series becomes:
√2, 2√2, 3√2, 4√2, …

We can see this is an Arithmetic Progression (A.P.) where —
👉 First term (a) = √2
👉 Common difference (d) = √2

So,
General term,
aₙ = a + (n - 1)d
= √2 + (n - 1)√2
= n√2

Formula for the Sum of n terms of an A.P.:

Sₙ = n/2 × [2a + (n - 1)d]

Substitute values:
S₁₀ = 10/2 × [2(√2) + (10 - 1)√2]
= 5 × [2√2 + 9√2]
= 5 × 11√2
= 55√2

Final Answer: (b) 55√2


Simplify square roots first to find a pattern.

Identify if the sequence is an A.P. (constant difference) or G.P. (constant ratio).

Use correct formula for the sum depending on the type of progression.

Important Formulas:

1️⃣ For A.P. →
  Tₙ = a + (n - 1)d
  Sₙ = n/2 × [2a + (n - 1)d]

2️⃣ For G.P. →
  Tₙ = arⁿ⁻¹
  Sₙ = a(1 - rⁿ) / (1 - r)

Learning Outcome:

After watching this video, you will learn:

How to simplify square roots and detect arithmetic patterns.

How to find nth term and sum of an arithmetic progression.

How to handle irrational (surd) sequences easily.

Step-by-step logical thinking to solve board and competitive exam problems.

Suitable for:

Class 9, Class 10, Class 11 Mathematics Students

WB Board / CBSE / ICSE / Assam Board / Bihar Board / Jharkhand Board

Competitive Exams: NTSE, Olympiad, SSC, Banking, PSC, NDA, CDS, Railway, etc.

Anyone preparing for concept-based mathematics.

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