CBSE Class 11 • 2024 • 4 Marks

Sequences and Series: Sum of Geometric Progression with Repeating Digits

Curated examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

Find the sum to $n$ terms of the sequence: $7, 77, 777, 7777, \dots$

Verified Solution & Marking Scheme

Factor out 7 and convert to 9s
$$S_n = 7 + 77 + 777 + \dots + (\text{to } n \text{ terms})$$ $$S_n = 7(1 + 11 + 111 + \dots + \text{to } n \text{ terms})$$ $$S_n = \frac{7}{9}(9 + 99 + 999 + \dots + \text{to } n \text{ terms})$$
Express each term as powers of 10 minus 1
$$S_n = \frac{7}{9} \left[ (10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1) \right]$$ $$S_n = \frac{7}{9} \left[ (10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots \text{ to } n \text{ times}) \right]$$
Apply the geometric series sum formula
For GP $10 + 10^2 + \dots + 10^n$: $a = 10, r = 10$. $$\text{Sum} = \frac{a(r^n - 1)}{r - 1} = \frac{10(10^n - 1)}{10 - 1} = \frac{10(10^n - 1)}{9}$$ $$\text{Subtracting } n: \quad S_n = \frac{7}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$$
Simplify the expression
$$S_n = \frac{70}{81}(10^n - 1) - \frac{7n}{9}$$
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