Question

How many terms of the sequence -8,-6 ,-4 ... Must be taken their sum is 70

Answer 1

Answer:

The sum of $14$ terms is $70$

Step-by-step explanation:

First term $a=-8$, common difference $d=2$,

sum to $n$ terms is $S_{n}=\frac{n}{2}[2a+(n-1)d]$.

We have to find $n$ such that $S_{n} =70$

$70=\frac{n}{2} [2(-8)+(n-1)2]$

   $=n[-8+n-1]=n[n-9]\\\\=n^{2} -9n$

Now solve the quadratic equation

$n^{2} -9n-70=0$

$n=\frac{-b}{2a}$±$\frac{\sqrt{b^{2}-4ac } }{2a}$

  $=\frac{9}{2}$±$\frac{\sqrt{81-4(1)(-70)} }{2}$

  $=\frac{9}{2}$±$\frac{\sqrt{361} }{2} }$

  $=\frac{9+19}{2}=14$

since number of terms is positive, negative value of $n$ need not be considered.

hence the answer

thank you