Let X and Y be age and no. of children.
values of y are given, then 5th leading difference will be zero.
Δ50 = (y – 1 )5 = y5 – 5y4 + 10y3 – 10y2 + 5y1 – y0 = 0 …. (i)
and the second equation can be obtained as, increasing the suffixes of each term of ‘Y’ by one, keeping the coefficients same; we get :
Δ5 = (y – 1)5 = y6 – 5y5+ 10y4 – 10y3 + 5y2 – y1 =0 …….(ii)
From equation (i)
y5 – 5y4 + 10y3 – 10y2+ 5y1 – y0 =0
ie., 98 – 5(88) + 10y3 – 10(65) + 5(55) – 50 = 0
98 – 440 + 10y3 – 650 + 275 – 50 = 0
So, by simplifying, 10y3 – 767 = 0
∴ y3 = 76.7 = 77 children.
From (ii)
i-e, y6 – 5y5 + 10y4 – 10y3 + 5y2 – y1 = 0;
Put y3= 76.7 we get, y6 – 5(98) + 10(88) – 10(76.7) + 5(65) – 55 = 0
i.e, y6 – 490 + 880 – 767 + 325 – 55 = 0
∴ y6 = 107 children.