Question

A number sequence has 100 elements. Any of its elements (except for the first and last element) is equal to the product of its neighbours. The product of the first 50 elements, just as the product of all the elements is 27. What is the sum of the first and the second element?

• A. 10
• B. 7
• C. 12
• D. 6

Answer 1

The correct option is C 12

Let first two numbers be x and y. So the sequence will be -

$x,y,\frac{y}{x},\frac{1}{x},\frac{1}{y},\frac{x}{y},x,y,.$ the sequence repeats after 6 numbers. Also the product of first six numbers is 1.

So now, product of first 50 numbers will be xy=27 (since the product of the other 48 will be 1) while product of first 100 numbers will be $\frac{y^2}{x}=27$ (since the product of the other 96 will be 1).
On solving we get y=9 and x=3.

So their sum = x+y =12.

Alternatively,

Look at the product, it is 27 which is nothing but $3^3$and the series is all about products. So the numbers in the series will be full of 3s. Look at the answer options for numbers which can be expressed as sum of two powers of 3. Only $12=3+3^2$ is the possibility.

Series is $3,3^2,3,\frac{1}{3},\frac{1}{3^2},\frac{1}{3},3,3^2.............$