Question

Express 225 as the sum of 15 odd numbers.

Answer 1

225=1+3+5+7+9+11+13+15+17+19+21+23+25+27+29


This can easy be find that given x² ,we can analyse that first x odd numbers sum is 225 .


Hope helped !

Answer 2

Answer:

225

Step-by-step explanation:

Taking a topic from Arithmetic Progression, we can conclude that the sum of, say n numbers, is:

$S_{n}=\frac{n}{2}(a+a_{n})$

Now $S_{n}[tex] is the sum. 'n' is the number of the given numbers. 'a' is the first term(first number) and [tex]a_{n}[tex] is the last term.</p><p>Now to find [tex]a_{n}$, we have a equation, which is:

$a_{n}=a+(n-1)d$

Here, 'd' is the common difference.

You can find 'd' by the equation:

$d=a_{2}-a_{1}$

Where $a_{1}$ is the first term and $a_{2}$ is the second term.

Now, coming back to your question, we know that the AP of the first 15 numbers is:

1,3,5,7......

Hence;

a=1

d=2 [using$(a_{2}-a_{1})$, we get(3-1=2)

now, we need to find $a_{15}$, aka the last term.

$a_{15}=a+(15-1)d$

⇒$a_{15}=1+(14)2$

∴$a_{15}=29$

Now, substituting, everything in the $S_{n}$ equation;

$S_{15}=\frac{15}{2}(1+29)$

⇒$S_{15}=\frac{15}{2}(30)$

⇒$S_{15}$=15×15

∴$S_{15}=225$

Using this method, you can add any number of numbers. Unlike the other answer, which requires you add all the numbers, one by one, this one will be quite helpful for adding a large amount of numbers which you cannot possibly add individualy. Also note that the numbers must be in an AP in order to use this formula.

Hope this helped!