Question

The sum of the series below is 10900 how many numbers n are in the series 19+20.5+22+23.5+...+181

Answer 1

109

Explanation:

`19,\ 20.5,\ 22,\ 23.5,\ ...,\ 181\\\\\text{It's the arithmetic sequence with:}\\\\a_1=19,\ a_n=181.\\\\\text{The sum of n terms of this sequence is equal to 10,900.}\\\\\text{the formula of a sum of n terms of an arithmetic sequence:}\\\\S_n=(a_1+a_n)/(2)\cdot n\\\\\text{Substitute:}\ S_n=10,900,\ a_1=19,\ a_n=181:\\\\(19+181)/(2)\cdot n=10,900\\\\(200)/(2)n=10,900\\\\100n=10,900\qquad\text{divide both sides by 100}\\\\n=109`

Answer 2

Consecutive terms differ by 1.5, so this is an arithmetic sequence given by

`\begin{cases}a_1=19\\a_n=a_(n-1)+1.5&\text{for }n>1\end{cases}`

So we have

`a_2=a_1+1.5`

`a_3=a_2+1.5=a_1+2\cdot1.5`

`a_4=a_3+1.5=a_1+3\cdot1.5`

and so on, up to

`a_n=a_1+1.5(n-1)=17.5+1.5n`

The sum of the first
`N` terms is 10,900:

`\displaystyle\sum_(n=1)^Na_n=\sum_(n=1)^N(17.5+1.5n)=17.5N+1.5\frac{N(N+1)}2=10,900`

`\implies17.5N+0.75N(N+1)=10,900`

`\implies0.75N^2+18.25N-10,900=0`

`\implies N=109` (we ignore the negative solution)

so there are 109 terms in the series.