Get an element by index in jQuery

I have an unordered list and the index of an li tag in that list. I have to get the li element by using that index and change its background color. Is this possible without looping the entire list? I mean, is there any method that could achieve this functionality?

Here is my code, which I believe would work...

<script type="text/javascript">
  var index = 3;
</script>

<ul>
    <li>India</li>
    <li>Indonesia</li>
    <li>China</li>
    <li>United States</li>
    <li>United Kingdom</li>
</ul>

<script type="text/javascript">
  // I want to change bgColor of selected li element
  $('ul li')[index].css({'background-color':'#343434'});

  // Or, I have seen a function in jQuery doc, which gives nothing to me
  $('ul li').get(index).css({'background-color':'#343434'});
</script>

$(...)[index]      // gives you the DOM element at index
$(...).get(index)  // gives you the DOM element at index
$(...).eq(index)   // gives you the jQuery object of element at index

DOM objects don't have css function, use the last...

$('ul li').eq(index).css({'background-color':'#343434'});

docs:

.get(index) Returns: Element

• Description: Retrieve the DOM elements matched by the jQuery object.
• See: https://api.jquery.com/get/

.eq(index) Returns: jQuery

• Description: Reduce the set of matched elements to the one at the specified index.
• See: https://api.jquery.com/eq/

You can use jQuery's .eq() method to get the element with a certain index.

$('ul li').eq(index).css({'background-color':'#343434'});

You can use the eq method or selector:

$('ul').find('li').eq(index).css({'background-color':'#343434'});

There is another way of getting an element by index in jQuery using CSS :nth-of-type pseudo-class:

<script>
    // css selector that describes what you need:
    // ul li:nth-of-type(3)
    var selector = 'ul li:nth-of-type(' + index + ')';
    $(selector).css({'background-color':'#343434'});
</script>

There are other selectors that you may use with jQuery to match any element that you need.