Mathematics: Sum of first p terms of an ap is 0 , if a is the first term , prove that the sum of next q terms is -aq(p+q)/p-1, p is not equal to 0

Sum of first p terms of an ap is 0 , if a is the

Ответ:

allyssaharrisooy50au

05.01.2022

Let

be the common difference between terms. Then the AP is

$\left\{\underbrace{a,a+d,a+2d,\ldots,a+(p-1)d}_{\text{first }p\text{ terms}},\underbrace{a+pd,a+(p+1)d,\ldots,a+(p+q+1)d}_{\text{next }q\text{ terms}}\right\}$

Because the first

terms add to zero, you have

$\displaystyle\sum_{n=1}^p (a+(n-1)d)=(a-d)\sum_{n=1}^p1+d\sum_{n=1}^pn=(a-d)p+d\frac{p(p+1)}2=0$

Solving for

yields $d=\dfrac{2a}{1-p}$ .

Now, the sum of the next

terms is

$\displaystyle\sum_{n=p+1}^{p+q}(a+(n-1)d)=\sum_{n=p+1}^{p+q}\left(a+(n-1)\frac{2a}{1-p}\right)$
$=\displaystyle\left(a-\frac{2a}{1-p}\right)\sum_{n=p+1}^{p+q}1+\frac{2a}{1-p}\sum_{n=p+1}^{p+q}n$
$=\displaystyle\left(a-\frac{2a}{1-p}\right)q+\frac{2a}{1-p}\left(\frac{q(2p+q+1)}2\right)$
$=\displaystyle\frac{aq(p+q)}{1-p}$
$=\displaystyle-\frac{aq(p+q)}{p-1}$

as desired.

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Ответ:

ccarwile01

29.07.2021

Mathematics

The answer is 37!

Step-by-step explanation:

22+1+14=37

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Sum of first p terms of an ap is 0 , if 'a' is the first term , prove that the sum of next q terms is -aq(p+q)/p-1, p is not equal to 0

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