Question

How many terms of AP - 6. - 11 by 2 .minus 5 are needed to be the sum -25​

Answer 1

Given:

The terms of a sequence are:

$-6,-\dfrac{11}{2},-5$

To find:

The number of terms whose sum is -25.

Solution:

We have, the given sequence:

$-6,-\dfrac{11}{2},-5$

Here, the first term is -6.

$-\dfrac{11}{2}-(-6)=-5.5+6$

$-\dfrac{11}{2}-(-6)=0.5$

Similarly,

$-5-(-\dfrac{11}{2})=-5+5.5$

$-5-(-\dfrac{11}{2})=0.5$

The difference between consecutive terms are same. So, the given sequence is an arithmetic sequence with common difference 0.5.

The sum of n terms of an arithmetic sequence is:

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

Where, a is the first term, d is the common difference.

Putting $S_n=-25,a=-6,d=0.5$, we get

$-25=\dfrac{n}{2}[2(-6)+(n-1)0.5]$

$-50=n[-12+0.5n-0.5]$

$-50=-12.5n+0.5n^2$

$0=0.5n^2-12.5n+50$

Splitting the middle term, we get

$0.5n^2-2.5n-10n+50=0$

$0.5n(n-5)-10(n-5)=0$

$(0.5n-10)(n-5)=0$

Using zero product property, we get

$(0.5n-10)=0$ and $(n-5)=0$

$n=\dfrac{10}{0.5}$ and $n=5$

$n=20$ and $n=5$

Therefore, the sum of either 5 or 20 terms is -25.