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W ciągu arytemetycznym a10=44 a14=56 wyznacz:a1,r,an i oblicz s20

Odpowiedź :

Kwiatkovaviz
a10 + 4r = a14
44 + 4r = 56
4r = 12
r = 3

an = a1 + (n-1)r
44 = a1 + (10-1)3
44 = a1 + 27
a1 + 27 = 44
a1 = 17

an = 17 + 3n - 3
an = 3n + 14

a20 = 17 + (20 - 1)3
a20 = 17 + 57
a20 = 74

S20 = 17 + 74 / 2 × 20 = 410
Vadizviz
W ciągu arytemetycznym a10=44 a14=56 wyznacz:a1,r,an i oblicz s20

a14= a10 + 4r
56=44+4r
r=3
an=a1+r(n-1)
44=a1+3*(10-1)
a1=17
an=14 +3n

a20 = 17 + 3*(20 - 1)
a20 = 17 + 57
a20 = 74

S_20 = 17 + 74 / 2*20 = 410
Janek191viz
a10 =44
a14= 56

a10 =a1 +9*r
a14 =a1 +13*r
-----------------
a14 -a10 = 13r - 9r = 4r =56-44 = 12
4r =12
r =3
a1 = a10 -9r = 44- 9*3 =44 -27 =17
a1 =17
an =17 +(n-1)*3 = 17 + 3*(n-1)
a20 =17 + 3*19 = 17+57 =74
S20 = [ a1 + a20] * 10 = [17 +74]*10 =91*10 =910
S20 = 910

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