Question

Wyznacz dla ciagu arytmetycznego (an):

liczbe n wyrazow, gdy Sn= 204, r= 6, an= 49

roznice r, gdy Sn= 518, a1= 50, n= 14

sume a5 + a6 + ... + a20, gdy an = 5n - 3

Answer 1

Sn = 204; r = 6; an = 49

-----------------------------

n = ?

mamy

an = a1 +(n -1)*r

a1 + (n-1)*6 = 49

a1 +6n - 6 = 49

a1 = 55 - 6n

===========

Sn = 0,5 *[a1 + an]*n

0,5*[55 - 6n + 49]*n = 204   / * 2

[104 - 6n]*n = 408

104n - 6n^2 = 408  / : 2

3 n^2 - 52 n + 204 = 0

------------------------------

delta = (-52)^2 - 4*3*204 = 2704 - 2448 = 256

p(delty) = 16

n = [ 52 - 16]/6 = 36/6 = 6

lub  n = [ 52 + 16]/6 = 68/6 = 11  1/3   < --- odpada ( nie jest naturalna)

Odp. n = 6

==============================================================

Sn = 518;  a1 = 50;  n = 14

-------------------------------

r = ?

 

an = a1 +(n-1)*r

an = 50 + 13r

Sn = [a1 + an]/2

0,5 *[ 50 + 50 + 13 r]*14= 518      / *2

[100 + 13r]*14 = 1036

1400 + 182 r = 1036

182 r = 1036 - 1400 = - 364 / : 182

r = -2

========

 

 

=======================

an = 5n - 3

zatem

a1 = 5*1 - 3 = 2

a2 = 5*2 - 3 = 7

a3 = 5*3 - 3 = 12

a4 = 5*4 - 3 = 17

S4 = a1+a2 +a3 + a4 = 2+7+12+17 = 38

zatem  r = 7 -2 = 12 - 7 = 5

a20 = a1 +19*r = 2 + 19*5 = 2 + 95 = 97

S20 = 0,5*[a1 +a20]*20 = 0,5*[2 +97]*20 = 10*99 = 990

zatem

a5 +a6 + a7 + .. + a19 + a20 = S20 - S4  = 990 - 38 = 952

Odp. 952

======================================================