What is the sum of the infinite series: 1 - 1/2 - 1/4 + 1/8 - 1/16 - 1/32 + 1/64 - 1/128 - 1/256 + ? Option 1: 6/7 Option 2: 27/32 Option 3: 2/7 Option 4: 5/8

Answer: 2/7. Solution: Looking at the signs: + - - + - - + - - + The pattern repeats every 3 terms, so we group accordingly. Group the terms: S = (1 - 1/2 - 1/4) + (1/8 - 1/16 - 1/32) + (1/64 - 1/128 - 1/256) + Calculate each group: Group 1: 1 - 1/2 - 1/4 Convert to same denominator: 4/4 - 2/4 - 1/4 = (4-2-1)/4 = 1/4 Group 2: 1/8 - 1/16 - 1/32 4/32 - 2/32 - 1/32 = (4-2-1)/32 = 1/32 Group 3: 1/64 - 1/128 - 1/256 4/256 - 2/256 - 1/256 = (4-2-1)/256 = 1/256 Rewrite as a new series: S = 1/4 + 1/32 + 1/256 + Notice that: 1/32 = 1/4 x 1/8 1/256 = 1/4 x 1/64 = 1/4 x (1/8)^2 So: S = 1/4(1 + 1/8 + 1/64 + ) The series inside parentheses is geometric with: - First term: a = 1 - Common ratio: r = 1/8 Using the infinite geometric series formula a/(1-r): Sum inside parentheses = 1/(1-1/8) = 1/(7/8) = 8/7 Final calculation: S = 1/4 x 8/7 = 8/28 = 2/7

Algebra

Progression & Series

Hard

What is the sum of the infinite series: $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots?$

$\frac{6}{7}$

$\frac{27}{32}$

$\frac{2}{7}$

$\frac{5}{8}$

✅ Correct Option: 3

Looking at the signs: $+ - - + - - + - - + \ldots$

The pattern repeats every 3 terms, so we group accordingly.

Group the terms:

$S = \left(1 - \dfrac{1}{2} - \dfrac{1}{4}\right) + \left(\dfrac{1}{8} - \dfrac{1}{16} - \dfrac{1}{32}\right) + \left(\dfrac{1}{64} - \dfrac{1}{128} - \dfrac{1}{256}\right) + \ldots$

Calculate each group:

Group 1: $1 - \dfrac{1}{2} - \dfrac{1}{4}$

Convert to same denominator:

$\dfrac{4}{4} - \dfrac{2}{4} - \dfrac{1}{4} = \dfrac{4-2-1}{4} = \dfrac{1}{4}$

Group 2: $\dfrac{1}{8} - \dfrac{1}{16} - \dfrac{1}{32}$

$\dfrac{4}{32} - \dfrac{2}{32} - \dfrac{1}{32} = \dfrac{4-2-1}{32} = \dfrac{1}{32}$

Group 3: $\dfrac{1}{64} - \dfrac{1}{128} - \dfrac{1}{256}$

$\dfrac{4}{256} - \dfrac{2}{256} - \dfrac{1}{256} = \dfrac{4-2-1}{256} = \dfrac{1}{256}$

Rewrite as a new series:

$S = \dfrac{1}{4} + \dfrac{1}{32} + \dfrac{1}{256} + \ldots$

Notice that:

$\dfrac{1}{32} = \dfrac{1}{4} \times \dfrac{1}{8}$

$\dfrac{1}{256} = \dfrac{1}{4} \times \dfrac{1}{64} = \dfrac{1}{4} \times \left(\dfrac{1}{8}\right)^2$

So:

$S = \dfrac{1}{4}\left(1 + \dfrac{1}{8} + \dfrac{1}{64} + \ldots\right)$

The series inside parentheses is geometric with:

First term: $a = 1$
Common ratio: $r = \dfrac{1}{8}$

Using the infinite geometric series formula $\dfrac{a}{1-r}$ :

$\text{Sum inside parentheses} = \dfrac{1}{1-\dfrac{1}{8}} = \dfrac{1}{\dfrac{7}{8}} = \dfrac{8}{7}$

Final calculation:

$S = \dfrac{1}{4} \times \dfrac{8}{7} = \dfrac{8}{28} = \dfrac{2}{7}$

2024: Slot 1

The sum of the first 3 terms of an AP is 6 and that of the last 3 terms is 16. If the AP has 13 terms, what is the Arithmetic mean of the three terms that are exactly at the middle of the progression?

2024: Slot 2

A boy agrees to work at the rate of one rupee on the first day, two rupees on the second day, four rupees on third day and

so on, thus getting on any day twice the amount he received just the day before. How many rupees will the boy earn in all, if

he started working on the $5^{th}$ of March and completed the work on the $28^{th}$ of March?

BBA-DBE Bangalore 2026

The 5th term of an arithmetic sequence is 20 and its 10th term is 35. The first term of a geometric sequence is the same as the first term of the arithmetic sequence mentioned above, while the 4th term of the geometric sequence is 1. What will be the sum to infinity of the arithmetic-geometric sequence whose $n$ -th term is given by $[a+(n-1)d]\times r^{n-1}$ , where $a$ is the first term of both the arithmetic and the geometric sequences, $d$ is the common difference of the arithmetic sequence, and $r$ is the common ratio of the geometric sequence?

BBA-DBE Bangalore 2026

A theatre has 30 rows of seats arranged in such a way that the number of seats in each row forms an arithmetic progression. The first row contains 18 seats, and each successive row has 3 more seats than the previous row. Due to renovation rules, only those rows whose seat count is divisible by 6 are reserved for premium booking. Find the total number of premium seats in the theatre.

BBA-DBE Bangalore 2026

Find the sum of the squares of the first twelve positive multiples of 8.

2024: Slot 2

Let $\dfrac{m}{n} = \dfrac{90^2+5 \times 90+4}{90^2+5\times 90+6} \times \dfrac{91^2+5\times 91+4}{91^2+5\times 91+6} \times .......\times \dfrac{98^2+5\times 98+4}{98^2+5\times 98+6}$ where m and n are co-primes.

What is the value of $|m - n|$ ?

What is the sum of the infinite series: 1−12−14+18−116−132+164−1128−1256+…?1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots?1−21​−41​+81​−161​−321​+641​−1281​−2561​+…?

Solution

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BBA-DBE Bangalore Past Year Questions (Topic-Wise)

What is the sum of the infinite series: $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \dots?$