Question

how many terms of an ap - 6 - 11 by 2 - 5 are needed to give the sum -25​

Answer 1

Answer:

$\sf The \:number \:of\: terms \:required\: to\: give\: the \:sum\: as -25\: is \:20 \:or\: 5 \:terms .$

Step-by-step explanation:

Here,

The first term (a) of the AP = -6

Common Difference , d = -11/2-(-6) = -11/2+6 = 1/2

We Know that ,

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

Here it is given that the sum is -25​

i.e S(n) = -25

So,

-25 = $\frac{n}{2} ( 2a+(n-1)d)$

- 50 = n( 2×-6+ (n-1)   1   )

                                  2

-50 = -12n + $\frac{n\²}{2} - \frac{n}{2}$

-50 ×2 = - 24+n² - n

-100 = -25n + n²

-100 = n² - 25n

n² - 25n -100 = 0

n² - 5n - 20n + 100

So , The solutions are ( x− 20 ) ( x-5)

Therefore , x= 20 and x = 5