Question

find the smallest number of terms which may be taken in order that the sum of the arithmetical series 325+350+375+.......may exceed 10000​

Answer 1

Answer: 19

Step-by-step explanation:

$a_n=a_1+d(n-1)\\a_n=325+25(n-1)\\.\quad =325+25n-25\\.\quad =25n+300\\\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\10000 =\dfrac{325+(25n+300)}{2}\cdot n\\\\\\10000=\dfrac{(25n+625)n}{2}\\\\\\20000=25n^2+625n\\\\\\0=25n^2+625n-20000\\\\\\0 = n^2+25n-800\qquad \text{(divided by 25)} \\\\\\n=18.4\qquad n=-43.4\qquad \text{(use quadratic formula to solve)}$

Since we can't have a negative number of terms,

n = -43.4 is an extraneous solution

--> n = 18.4 is the only valid solution

In order to EXCEED 10000, n must be GREATER THAN 18.4

The first integer greater than 18.4 is ....

                                        19

Answer 2

Answer is 19;

Problem
a1=325 , d=25 , S19=?
Result
S19=10450
Explanation
To find S19 we use formula
Sn=n2⋅(2a1+(n−1)⋅d)
In this example we have a1=325 , d=25 , n=19. After substituting these values into the above equation, we obtain:
Sn19=n2⋅(2a1+(n−1)⋅d)=192⋅(2⋅325+(19−1)⋅25)=192⋅(650+18⋅25)=192⋅(650+450)=192⋅1100=10450