Question

The first and second terms of a linear sequence (A.p) are 3 and 8 respectively. Please determine the least Number of terms of Ap that must be added so that the sum is greater than 250​

Answer 1

Answer:

To rewrite your question, you are looking for the smallest    that fulfills the inequality:

∑=1>250

First, we must find the sequence    in explicit form. We can think of it as a line we are trying to get the slope-intercept form of. We have the points (1, 3) and (2, 8). Therefore, the slope is  8−32−1=51=5 . Now we just need the y-intercept. We can find it through substitution:

=5+

In order to solve for   , we can plug in one of the points that we were already given. I will choose the point (1, 3). Note that this point just means that when  =1 ,  =3 .

3=5(1)+

3=5+

=−2

Now for the original question.

∑=1(5−2)>250

((5(1)−2)+(5()−2)2)>250

(5+12)>250

522+12>250

2+15>100

2+15+1100>100+1100

(+110)2>100+1100

(+110)2>10001100

+110>±10001100‾‾‾‾‾‾‾√

+110>±11010001‾‾‾‾‾‾√

>−110±11010001‾‾‾‾‾‾√

Since we are looking for the smallest value, we will subtract for the plus or minus sign.

>−110−11010001‾‾‾‾‾‾√≈−10

Since that is negative, we will have to add instead.

>−110+11010001‾‾‾‾‾‾√≈9.9

Therefore, the smallest integer    that makes sense and satisfies the inequality is 10.

So, the first 10 numbers must be added.

Honestly, it would have been quicker just to brute-force this.

Step-by-step explanation:

Have a good day