Find the value of n given First term is a=-16, Common difference d=8, the sum of first n terms is 600?

joguejaseg
2022-09-21
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Jacey Humphrey

Answered 2022-09-22
Author has **7** answers

This is an Arithmetic sequence

The sum of the first n terms of this sequence is given by.

${S}_{n}=\frac{n}{2}[2a+(n-1)d]$

where a is the first term and d , the common difference.

here a = -16 , d = 8 and require to solve for n.

hence : $\frac{n}{2}[(2\times -16)+8(n-1)]=600$

$\frac{n}{2}[-32+8n-8]=600\Rightarrow \frac{n}{2}(8n-40)=600$

distributing gives : $4{n}^{2}-20n-600=0$

Equated to zero since this is a quadratic equation.

$\Rightarrow 4({n}^{2}-5n-150)=0$

$\Rightarrow 4(n+10)(n-15)=0\Rightarrow n=-10\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}n=15$

but n > 0 hence n = 15

The sum of the first n terms of this sequence is given by.

${S}_{n}=\frac{n}{2}[2a+(n-1)d]$

where a is the first term and d , the common difference.

here a = -16 , d = 8 and require to solve for n.

hence : $\frac{n}{2}[(2\times -16)+8(n-1)]=600$

$\frac{n}{2}[-32+8n-8]=600\Rightarrow \frac{n}{2}(8n-40)=600$

distributing gives : $4{n}^{2}-20n-600=0$

Equated to zero since this is a quadratic equation.

$\Rightarrow 4({n}^{2}-5n-150)=0$

$\Rightarrow 4(n+10)(n-15)=0\Rightarrow n=-10\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}n=15$

but n > 0 hence n = 15

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