Question

Question #935ae

Answer 1

`1200`

Explanation:

The sum of an arithmetic sequence is:

`n/2(2a+(n-1)d`

We need to find first term `a` and common difference `d`:

The nth term is:

`a+(n-1)d`

4th term:

`a + (4-1)d=84`

`a+3d=84`

10th term:

`a+(10-1)d=60`

`a+9d=60`

Solving simultaneously:

`a+3d=84color(white)(8888)[ 1 ]`
`a+9d=60color(white)(8888)[ 2 ]`

Subtracting [ 2 ] from [ 1 ]

`-6d=24=>d=>-4`

`a+3d=84color(white)(8888)[ 1 ]`

Substituting in [ 1 ]

`a+3(-4)=84=>a=96`

Because the terms are decreasing, they will start to become negative. We need the maximum possible sum, so we need to find the last positive term. Using:

`a+(n-1)d`

`96+(n-1)(-4)=96-4n+4=100-4n`

`100-4n=0=>n=25`

25th term is:

`96+24(-4)=0`

26th term is:

`96+25(-4)=-4`

So we could use the sum to the 24th term or the 25th, since the 25th term is zero it makes no difference to the sum. we will use the 24th though.

`24/2(2(96)+(23(-4))=(24)/2(100)=1200`