Так как среднее значение 3 чисел (СРЗНАЧ(A1:A3)) равно 3, то сумма значений в этих же ячейках = 3*3=9
Значит значение ячейки A4 = 10-9 = 1
Можно цикл запустить наооборот:
for i:=n downto 1 do
Writeln(x[i]);
X=R2C2=RC[-4]
y=R2C3=RC[-3]
z=R2C4=RC[-2]
1. =3,4*RC[-3]+RC[-2]
2. =(R[-1]C[-3]+R[-1]C[-2])*R[-1]C[-1]
3. =0,8*R[-2]C[-3]+0,9*R[-2]C[-2]-R[-2]C[-3]*R[-2]C[-2]
4. =(R[-3]C[-3]+R[-3]C[-1])*R[-3]C[-2]+0,1*R[-3]C[-3]
5. =(R[-4]C[-3]-R[-4]C[-1])*R[-4]C[-1]+R[-4]C[-2]*R[-4]C[-1]
6. =6*((R[-5]C[-3])^3-5*R[-5]C[-2]/6)/((R[-5]C[-3])^3-6*R[-5]C[-2])
7. =(15*(R[-6]C[-3])^2-7*R[-6]C[-2]/12)/(18*R[-6]C[-2]+(R[-6]C[-3])^2)
8. =(40*(R[-7]C[-2])^3+4*R[-7]C[-3]/9)/(6*(R[-7]C[-3])^2-18*R[-7]C[-3]*R[-7]C[-2])
9. =(((5*(R[-8]C[-3])^(3*R[-8]C[-2])-2))/(10*R[-8]C[-3]*R[-8]C[-2]))/((12*R[-8]C[-3]+R[-8]C[-2])/(3*R[-8]C[-3]-(R[-8]C[-2])^(5+2*R[-8]C[-3])))
<span>10. =((5*(R[-9]C[-3])^(3+R[-9]C[-2])-2)/(10-R[-9]C[-3]*R[-9]C[-2]))/(12*R[-9]C[-3]*R[-9]C[-2]/(3*R[-9]C[-3]+(R[-9]C[-2])^2*R[-9]C[-3])))
</span>
Uses crt;
var i,N:integer;
s:real;
begin
s:=0;
write('вв N');
read(N);
for i:=1 to N do
s:=s+sqrt(i);
write('otvet:', s);
readkey
end.