Question

Plz help me with this question.....​

Answer 1

See Below.

Explanation:

The sum of an A.P. is given by:

`\displaystyle S_n=(n)/(2)(a+x_n)`

Where n is the number of terms, a is the initial term, and xₙ is the last term.

Therefore:

`\displaystyle S_(30)=(30)/(2)\Big(a+x_(30)\Big)=15(a+x_(30))`

And likewise:

`\displaystyle S_(20)=(20)/(2)\Big(a+x_(20)\Big)=10(a+x_(20))\\\\ S_(10)=(10)/(2)\Big(a+x_(10)\Big)=5(a+x_(10))`

Substitute:

`15(a + x_(30)) = 3(10 (a + x_(20)) - 5(a + x_(10) ))`

Distribute:

`15a+15x_(30)=30a+30x_(20)-15a-15x_(10)`

Simplify:

`15x_(30)=30x_(20)-15x_(10)`

Simplify:

`x_(30)=2x_(20)-x_(10)`

The nth term for an A.P. is:

`x_n=a+d(n-1)`

Where a is the initial term and d is the common difference.

So, it follows that:

`x_(30)=a+d(30-1)=29d+a\\ x_(20)=a+d(20-1)=19d+a\\ x_(10)=a+d(10-1)=9d+a`

Therefore:

`29d+a=2(19d+a)-9d-a`

Which follows that:

`29d+a=38d+2a-9d-a\\ \\ 29d=29d \\\\ d\stackrel{\checkmark}{=}d`

QED.