Find Missing terms in interpolation table

x	f(x)
1961	200
1962	220
1963	260
1964	?
1965	350
1966	?
1967	430

Solution:
The value of table for `x` and `y`

x	1961	1962	1963	1964	1965	1966	1967
y	200	220	260	?	350	?	430

Here we are given 5 values of y. We assume that y is polynomial of degree 4.
Hence `Delta^5 y_k=0`

`(E-1)^5 y_k=0`

`(E^5-5E^4+10E^3-10E^2+5E-1) y_k=0`

taking `k=0`

`y_5-5y_4+10y_3-10y_2+5y_1-y_0=0`

`:.b-5(350)+10(a)-10(260)+5(220)-200=0`

`:.b+10a-3450=0`

`:.b+10a=3450`

taking `k=1`

`y_6-5y_5+10y_4-10y_3+5y_2-y_1=0`

`:.430-5(b)+10(350)-10(a)+5(260)-220=0`

`:.-5b-10a+5010=0`

`:.-5b-10a=-5010`

Now solving this 2 equations using substitution method
`b+10a=3450`

and `-5b-10a=-5010`

`-5(b+2a)=-5 * 1002`

`b+2a=1002`

Suppose,
`10a+b=3450 ->(1)`

and `2a+b=1002 ->(2)`

Taking equation `(1)`, we have

`10a+b=3450`

`=>b=-10a+3450 ->(3)`

Putting `b=-10a+3450` in equation `(2)`, we get

`2a+b=1002`

`2a+(-10a+3450)=1002`

`=>2a-10a+3450=1002`

`=>-8a+3450=1002`

`=>-8a=1002-3450`

`=>-8a=-2448`

`=>a=306 ->(4)`

Now, Putting `a=306` in equation `(3)`, we get

`b=-10a+3450`

`=>b=-10(306)+3450`

`=>b=-3060+3450`

`=>b=390`

`:.b=390" and "a=306`

---

---