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int[] numbers = { 1, 4, 7, 10 };
int product = numbers.Aggregate(1, (interim, next) => interim * next);
Console.WriteLine(product); // output: 280

What I have tried:

I am unable to understand how this code works
Updated 9-Dec-19 11:01am
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Solution 2

It's a Linq Method - Aggregate - which takes two parameters: a starting value - 1, and a lambda expression - (interim, next) => interim * next which explains what to return.
The lambda comes in two parts, separated by "=>":
1) The first part names the parameters as interim (the result so far) and next (the value for this element.
2) The second part is is the expression which generates the total for each element as it loops through them.

Think of it as a "raw" loop:
int product = 1;
foreach (int next in numbers)
    product = product * next;
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Solution 3

Here's the definition for the extension method.

public static TAccumulate Aggregate<TSource, TAccumulate>(this IEnumerable<TSource> source, TAccumulate seed, Func<TAccumulate, TSource, TAccumulate> func);

As Griff says in his solution, it's equivalent to a for loop. It can be confusing, the first parameter is a seed, it does not reference the first item in the array, it's an additional value. So, if it was 2 instead of 1, the result would be 560. I prefer to name the parameters in a similar way to those given in the definition and use 'item' for an item of the enumerable.
int[] numbers = { 1, 4, 7, 10 };
int product = numbers.Aggregate(seed:2,(accumulator, item) => accumulator * item);
Console.WriteLine(product); // output: 560

You don't actually need to set the seed value in your example. There is an overload of the method that uses the first value of the enumerable as the seed.
int[] numbers = { 1, 4, 7, 10 };
int product = numbers.Aggregate((accumulator, item) => accumulator * item);

An interesting point about the seed is that it can be a different Type to the enumerable Type but it's of the same Type that's returned from the method. There's another overload of the method that lets you use the result of the aggregation as the parameter of a function. So you can do something like this. Define the seed as a double and return a double from the function.
var intCollection = new int[] { 1,2,3,4 };
double squareRoot10 = intCollection.Aggregate(0.00D, 
(accumulator, item) => accumulator += item,
accumulator=> Math.Sqrt(accumulator));

A more practical example would be to construct your own accumulator Type with methods to process the 'items' and to return the result of the accumulation.

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Solution 1

This code is using the aggregate function[^] to return the product of elements of specified array.

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