Question

intr-o progresie aritmetica cu ratia 4 avem a6=-13 calculati a10 .

Answer 1

$\it a_{10} = a_6+4r \Longrightarrow a_{10} = -13 +4\cdot4=-13+16 = 3$

Answer 2

r=4

a6=-13

a10=?

a6=a1+5r

-13=a1+5*4

-13=a1+20

-13-a1-20=0

-a1-33=0

-a1=33 /(-1)

a1=-33

a10=a1+9r= -33+9*4=-33+36=3