Question

How many terms are required to make a sum greater than 600 when the 8th term is 11 and the 15 term is 21?

Answer 1

29 terms

Explanation:

`T_n=a+(n-1)d`
where:
`T_n` is a term
a= the first number in the sequence
n= represents the nth term
d= difference between two adjacent terms ie `T_(n-1)-T_n`

So the 8th term is 11
ie `T_8=11=a+(8-1)d`

`11=a+7d` --- (1)

And the 15th term is 21
ie `T_15=21=a+(15-1)d`

`21=a+14d` --- (2)

(2) - (1)
`10=7d`
`d=10/7` --- (3)

Sub (3) into (1)

`11=a+7times10/7`
`11=a+10`
`a=1`

The sum of the terms is given by:
`S_n=n/2[2a+(n-1)d]`
where
n= total number of terms
a= first term
d= difference

For `S_n>600`

`n/2[2(1)+(n-1)10/7]>600`

`n[2+10/7n-10/7]>1200`

`n[4/7+10/7n]>1200`

`n(4+10n)>8400`

`n(2+5n)>4200`

`2n+5n^2>4200`

`5n^2+2n-4200>0`

Using the quadratic formula:

`n=(-b+-sqrt(b^2-4ac))/(2a)`

`n=(-2+-sqrt(2^2-4(5)(-4200)))/(2(5))`

`n=(-2+-sqrt84004)/10`

`n=(-2+-2sqrt21001)/10`

`n=(-1+-sqrt21001)/5`

`n= 28.78 "or" n=-29.18`

Since n= number of terms, then `n>0`

`n=28.78` ie 29 terms are required