IPMAT Indore 2019 (MCQ) - There are numbers a_1, a_2, a_3, , a_n each of them being +1 or -1. If it is known that a_1 a_2 + a_2 a_3 + a_3 a_4 + a_n-1 a_n + a_n a_1 = 0 then | PYQs + Solutions

IPMAT Indore 2019

Algebra

Progression & Series

Medium

There are numbers $a_1, a_2, a_3, \ldots, a_n$ each of them being $+1$ or $-1$ . If it is known that $a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots a_{n-1} a_n + a_n a_1 = 0$ then

n

is a multiple of 2 but not a multiple of 4

n

is a multiple of 3

n

can be any multiple of 4

The only possible value of

n

is 4

✅ Correct Option: 3

Given:

\newline

Each

aᵢ ∈ {+1, -1} for i = 1, 2, ..., n

a₁a₂ + a₂a₃ + a₃a₄ + ... + aₙ₋₁aₙ + aₙa₁ = 0

Let's denote

S = a₁a₂ + a₂a₃ + a₃a₄ + ... + aₙ₋₁aₙ + aₙa₁.

Since each

aᵢ

is either

+1

-1

, each product

aᵢaᵢ₊₁

is also either

+1

-1

\newline

\newline

For

S = 0

, the sum of n terms (each being ±1) to equal zero.

This means:

Number of

(+1)

terms = Number of

(-1)

terms

Therefore,

n

must be even

Now let's consider what values of

n

are possible.

For

n = 2:

We have

a₁a₂ + a₂a₁ = 2a₁a₂ = 0

, which is impossible since

a₁a₂ = ±1

For

n = 4:

We have

a₁a₂ + a₂a₃ + a₃a₄ + a₄a₁ = 0

This is possible. For example:

a₁ = 1, a₂ = 1, a₃ = -1, a₄ = -1

Then:

(1)(1) + (1)(-1) + (-1)(-1) + (-1)(1) = 1 - 1 + 1 - 1 = 0 ✓

The key insight is that we're looking at a cyclic sum where consecutive elements are multiplied. Since each product is

±1

and we need the sum to be

0

, we need exactly

n/2

products to be

+1

and

n/2

products to be

-1

After working through the algebra and constraints, it turns out that n must be a multiple of

4

for the equation to have solutions.

For

n ≡ 2

(mod

4

), it can be shown that no solution exists due to the cyclic nature of the constraint.

Therefore, the answer is:

n

can be any multiple of

4

IPMAT Indore 2019 (MCQ) PYQs

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