Let the diameter be x and area be y.
To find y when x = 82, use Newton's forward interpolation form
∴ x0 + nh = 82 ⇒ 80 + n(5) ⇒ 82 - 80 = 2
⇒ n = \(\frac{2}{5}\) = 0.4
The difference table is

∴ y(82) = y0 + \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ \(\frac { n(n+1)(n+2)(n-3) }{ 4! } ​​{ \triangle }^{ 4 }{ y }_{ 0 }​​​​​​​​​​​​​​\)

+ \(\frac{(0.4)(0.4-1)(0.4-2)}{6}\)(-2)

= 5026 + 259.2 + (0.4) (-0.6) (20) + \(\frac{(0.4)(-0.6)(-1.6)(-1)}{3}\) + \(\frac{(0.4)(-0.6)(-1.6)(-2.6)}{3}\)
= 5026 + 259.2 - 4.8 - 0.128 - 0.1664
= 5280.10
∴ When the diameter is 82, area of circle is 5280.1 (≅ 5281)
To find y when x = 91, use Newton's backward interpolation formula.
∴ x_n + nh = 91 ⇒ 100 + n(5) = 91
⇒ 5n = 91 - 100
⇒ 5n = -9 ⇒ n = \(\frac{-9}{5}\) = 1.8
Newton's backward interpolation formula is
y_(x=xn+nh) = yn + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\)
y(x = 91) = 7854 - 1.8 (766) +

\(+\frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{3!}\)(4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + (-1.8) (-0.8) (20) + (-1.8) (-0.8) (0.2) + \(\frac{(-1.8)(-0.8)(0.2)(1.2)}{6}\) (4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + 28.8 + 0.288 + 0.2304
⇒ y_{(x = 91)} = 6504.5
Hence when the diameter is 91, area is 6504.5 ≅ 6504

Since x = 43 lies in the beginning of the table, use.
Newton's forward interpolation formula.
∴ x₀+ nh = 43 ⇒ 40 + n(10) = 43 ⇒ 10n
=43 - 40 = 3
⇒ n = \(\frac{3}{10}\) = 0.3
Newton's forward interpolation formula is
y_(x=x0+nh) = y₀\(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }​​​​​​​​​​​​​​\)
The difference table is

= 184 + 6 + (0.3)(-0.7)
= 184 + 6 - 0.21
= 189.79
∴ when x = 43, y = 189.79.
To find y when x = 84, use Newton's backward interpolation formula
∴ x_n + nh = 84
90 + n(10) = 84 ⇒ 10n = 84 - 90 = -6
⇒ n = \(\frac{-6}{10}\) = -6
Newton's backward interpolation formula is
∴ y_(x=xn+nh)= y_n + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\) 

= 304 - 16.8 + (-0.6) (0.4)
= 304 - 16.8 - 0.24
= 304 - 17.04 = 286.96
∴ when x = 84, y = 286.96

Since only four values are given,
(E -1)4 u₀ = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) u₀
= u₄ - 4 u₃ + 6 u₂ - 4 u₁ + u₀
⇒ 385 - 4(u₃) + 6(520) - 4 (556) + 560 =0
⇒ 385 - 4u₃ + 3120 - 2224 + 560 = 0
⇒ 1841 - 4u₃ = 0
⇒ 1841 = 4u₃
⇒ u₃ = \(\frac{1841}{4}\) = 460.25
∴ u₃ = 460.25

Given
The difference table is

To findy when x = x ⇒ x₀+ nh = x ⇒ 1 + n (1) = x ⇒ n = x-1
Newton's forward interpolation formula is
y_(x=x) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ .....
y_(x=x) = 1 + (x-1)(-2) + \(\frac { (x-1)(x-2) }{ 2 } (4)+\frac { (x-1)(x-2)(x-3) }{ 6 } (-8)+\frac { (x-1)(x-2)(x-3)(x-4) }{ 24 } (16)\)
⇒ y = 1 - 2x + 2(x² - 3x + 2) - \(\frac{4}{3}\) (x - 1) (x - 2) (x - 3) + \(\frac{2}{3}\) (x - 1) (x - 2) (x - 3) (x -4)
⇒ y = 3 - 2x + 2x² - 6x + 4 - \(\frac{4}{3}\) [(x² - 3x + 2) (x- 3)] + \(\frac{2}{3}\) [(x² - 3x + 2)(x² - 7x + 12)]
⇒ y = 2x² - 8x+ 7- \(\frac{4}{3}\) [x³- 3x²+ 2x- 3x² + 9x- 6] + \(\frac{2}{3}\) [x⁴ - 3x³ + 2x² - 7x³ + 21x² - 14x + 12x² - 36x + 24]
⇒ y = 2x²-8x+7- \(\frac{4}{3}\) x³ + 4x² - \(\frac{8}{3}\) x + 4x² - 12x + 8 + \(\frac { { 12x }^{ 4 } }{ 3 } -\frac { 20 }{ 3 } { x }^{ 3 }+\frac { 70 }{ 3 } { x }^{ 2 }-\frac { 100x }{ 3 } +\frac { 48 }{ 3 } \)
⇒ y = \(\frac{2}{3}\) x⁴ + x³ \(\left( \frac { -4 }{ 3 } \frac { -20 }{ 3 } \right) \) + x²\(\left( 2+4+4+\frac { 70 }{ 3 } \right) \) + x \(\left( -8-\frac { 8 }{ 3 } -12-\frac { 100 }{ 3 } \right) \) + 31
⇒ y =  \(\frac{2}{3}\) x⁴ - 8x³ + \(\frac{100}{3}\) x² - 56x + 31 which is the required polynomial.

Given

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

Given

Since we have to find f(0.5) which is at the beginning of the table, use Newton's forward interpolation formula.
x_n + nh = 0.5 ⇒ -1 + n(1) = 0.5
⇒ n = 0.5 + 1 = 1.5
∴ y_(0.5) = \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)
The difference table is

∴ y(0.5) = 202+\(\frac { 1.5 }{ 1! } (-27)+\frac { (1.5)(1.5-1) }{ 2! } (-66)+\frac { (1.5)(1.5-1)(1.5-2) }{ 3! } (132)\)
= 202 - 40.5 + (1.5) (.5) (-33) + \(\frac { (1.5)(.5)(-0.5) }{ 6(132) } \) (132)
= 202 - 40.5 - 24.75 - 8.25
= 202 -73.5
= 128.5
Hence f(0.5) = 128.5

Let the missing entries be y₂ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero
⇒ (E - 1)⁴y_k = 0
⇒ (E₄- 4E₃ + 6E₂ - 4E + 1) y_k = 0  (1)
Put k = 0 in (1) we get,
y₄ - 4y₃ + 6y₂ - 4y₁ + y₀ = 0
y₄ - 4(18) + 6y₂ - 4(11) + 7 = 0
⇒ y₄ - 72 + 6y₂ - 44 + 7 = 0
⇒ y₄ + 6y₂ = 109    (2)
Put k = 1 in (1) we get,
(E₄- 4E₃ + 6E₂ - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ +y₁ = 0
⇒ 32 - 4 (y₄) + 6 (18) - 4y₂ + 11 = 0
32 - 4y₄ + 108 - 4y₂ + 11 = 0
⇒ -4y₄ - 4y₂ + 151 = 0

Adding,           
⇒ y₂ = 14.25
Substituting y₂ = 14.25 in (2) we get,
y₄ + 6 (14.25) = 109
⇒ y₄ + 85.5 = 109
⇒ y₄ = 109 - 58.5
⇒ y₄ = 23.5

Using interpolation, find the value of f(x) when x =1
Here the intervals are unequal. By Lagrange's interpolation formula, we have
x₀=3, x₁ = 7, x₂ = 11, x₃ = 19
y₀ = 42, y₁ = 43, y₂ = 47, y₃ = 60 and x = 15
∴ Y =f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (15-7)(15-11)(15-19) }{ (3-7)(3-11)(3-19) } \times 42+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 43+\frac { (15-3)(15-7)(15-19) }{ (11-3)(11-7)(11-19) } \times 47+\frac { (15-3)(15-7)(15-11) }{ (19-3)(19-7)(19-11) } \times 60\)

= \(\frac{21}{2}-43+70.5+15\)
= 10.5 - 43 + 70.5 + 15
y= 53
Hence when x = 15, f(x) = 53.

Given:

Here, the intervals are unequal:
∴ By Lagrange's interpolation formula, we have
x₀ = 1982, x₁ = 1983, x₂ = 1984, x₃ = 1986
y₀ = 150, y₁ = 235, y₂ = 365, y₃ = 525 and x = 1985
=\(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)  \(\frac { (1985-1983)(1985-1984)(1985-1986) }{ (1982-1983)(1982-1984)(1982-1986) } \times 150+\frac { (1985-1982)(1985-1984)(1985-1986) }{ (1983-1982)(1983-1984)(1983-1986) } \times 235+\frac { (1985-1982)(1985-1983)(1985-1986) }{ (1984-1982)(1984-1983)(1984-1986) } \times 365+\frac { (198 5-1982)(1985-1983)(1985-1984) }{ (1986-1982)(1986-1983)(1986-1984) } \times 525\)

\(=\frac { (2)(1)(-1) }{ (-1)(-2)(-4) } \times 150+\frac { (3)(1)(-1) }{ (1)(-1)(-3) } \times \) 
= 37.5 + 235 + 547.5 + 131.25
= 481.25

Here the intervals are unequal
∴ By Lagrange'sinterpolation formula, we have
x₀ = 15, x₁ = 25, x₂ = 30, x₃ = 35
y₀= 36, y₁ = 40, y₂ = 45, y₃ = 38, and x = 26
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (26-25)(26-30)(26-35) }{ (15-25)(15-30)(15-35) } (36)+\frac { (26-15)(26-1530)(26-35) }{ (25-15)(5-30)(25-35) } (40)+\frac { 26-15)(26-25)(26-35) }{ (30-15)(30-30)(30-35) } (40)+\frac { (26-15)(26-25)(26-30) }{ (35-15)(35-25)(35-30) } (38)\)
= \(\frac { (1)(-4)(-9) }{ (-10)(-15)(-20) } (36)+\frac { (n)(-4)(-9) }{ (10)(-5)(-10) } (40)+\frac { (11)(1)(-9) }{ (15)(5)(-5) } (45)+\frac { (9)(1)(-4) }{ (-20)(10)(5) } (38)\)
= \(-\frac { (36)(36) }{ (150)(20) } +\frac { (44)(9)(40) }{ 500 } +\frac { 99(45) }{ (15)(25) } -\frac { (36)(38) }{ (200)(15) } \)
= \(\frac { 1296 }{ 3000 } +\frac { 15840 }{ 500 } +\frac { 4455 }{ 375 } -\frac { 1368 }{ 1000 } \)
= -0.432 + 31.68 + 11.88 - 1.368
= 41.76.
Hence the number of workers getting income not exceeding Rs. 26 per month is 42.

Given

 Here the intervals are unequal.
∴ By Lagranges interpolation formula, we have
x₀ = 1974, x₁ = 1978, x₂ = 1982, x₃ = 1990
y₀ = 25, y₁ = 60, y₂ = 80, y₃ = 170 and x = 1986.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
\(\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times25+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times60+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times80+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times 70+\frac { (1986-1974)(1986-1982)(1986-1990) }{ (1990-1974)(1990-1978)(1990-1982) } \times 170\)

= 6.25 - 60 + 120 + 42.5
y = 108.75

Given

 Find f(2.8)
Since the required value 2.8 is at the end of the table, apply Newton's backward interpolation formula.
x_n + nh = 2.8 ⇒ 3 + n (1) = 2.8
⇒  n = 2.8 - 3 = -0.2
The difference table is

Newton's backward interpolation formula is
y(x = x_n + nh) = \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
⇒ y(2.8) = 34 + (-0.2) (23) + \(\frac { (-0.2)(-0.2+1) }{ 2 } (14)+\frac { (0.2)(-0.2+1)(-0.2+2) }{ 6 } \)(16)
⇒ y(2.8) = 34 - 4.6 + (-0.2) (0.8) (7) + (-0.2) (0.8) (1.8)
⇒ y(2.8) = 34 - 4.6 - 1.12- 0.288
⇒ y(2.8) = 27.992

Let the percentage of lead be x and temperature be y.
Given

Find y when x = 84
Since the temperature required is at the end of the table we apply Newton's backward interpolation formula.
∴ x_n + nh = 84 ⇒ 90 + n(10) = 84
⇒ 10n = 84 - 90 = -6
[∵ h = 10 & x_n = 90]
⇒ n = \(\frac{-6}{10}\) = -0.6
The difference table is

∴ The Newton's backward interpolation formula is
y_(y=xn+nh) = \(\frac { n }{ 1! } \Delta { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3) }{ 4! } { \triangledown }^{ 4 }{ y }_{ n }+\frac { n(n+1)(n+2)(n+3)(n+4) }{ 5! } { \triangledown }^{ 5 }{ y }_{ n }\)
∴ y(84) = 304 + (-0.6) (28) + \(\frac { (-0.6)(-0.6+1) }{ 2 } (2)+0+0+\frac { (-0.6)(-0.6+1)(-0.6+3)(-0.6+4) }{ 5! } (4)\) [∵ ∇³ & ∇⁴ are zero]
⇒ y(84) = 304 - 16.8+ (-0.6) (0.4) + \(\frac { (-0.6)(0.4)(1.4)(2.4)(3.4) }{ 120 } (4)\)
⇒ y(84) = 304 -16.8 - 0.24 - 0.09139
⇒ y(84) = 286.86
Hence, the melting point of the alloy containing 84 percent lead is 286.86°C.

To find y, when x = 32.
Since 32 is near the beginning of the table, we use the Newton's forward interpolation formula
\(\frac { n }{ 1! } \Delta { y }_{ 0 }+\frac { n(n+1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }\)
∴ x₀ + nh = 32 ⇒ 30 + n(5) = 32 [∵ h = 5]
⇒ 5n = 2
⇒ n = \(\frac{2}{5}\) = 0.4
The difference table is

∴ y_(x=32) = 15.9 + 0.4(-1) + \(\frac { (0.4)(0.4-1)(0.2) }{ 2 } +\frac { (0.4)(0.4-1)(0.4-2)(-0.2) }{ 3! } +\frac { (0.4)(0.4-1)(0.4-2)(0.4-3)(0.2) }{ 4! } \)
= 15.9 - 0.4 + (0.2) (-0.6) (0.2) + \(\frac { (0.4)(-0.6)(-1.6)(-0.2) }{ 6 } +\frac { (0.4)(-0.6)(-1.6)(-2.6)(0.2) }{ 24 } \)
= 15.9 - 0.4 - 0.24 - 0.0128 - 0.002476
= 15.245
Hence, when x = 32, f(x) = 15.45

Given

 This can be rewritten as

Since we have to find below 70, use Newton's backward interpolation formula
∴ x_n + nh = 70 ⇒ 99 + n(20) = 70
⇒ 20n = 70 - 99 = -29
⇒ n = \(\frac{-29}{20}\) = -1.45
∴ y₇₀ = y_n + \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
\(=235-1.45(17)+\frac { (-1.45)(-1.45+1) }{ 2 } (-33)+\frac { (-1.45)(-1.45+2)(-18) }{ 6 } \)
\(=235-24.65+\frac { (-1.45)(-.45) }{ 2 } (-33)+\) (-1.45)(-0.45)(0.55)(-3)
= 235 - 24.65 - 12.375 - 1.125
= 196
Hence, the number of students who have scored below 70 are 196 (app).

Given

Since the population in the year 1955 is asked, which is near the beginning of the table, we have to follow Newton's forward interpolation formula.
x₀ + nh = 1955 ⇒ 1951 + n(10) = 1955
⇒ 10n = 1955 - 1951 = 4
⇒ \(\frac{4}{10}\) = 0.4
The difference table is

Newton's forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\({ y }_{ (x=1955) }=35+0.4(7)+\frac { (0.4)(0.4-1) }{ 2 } (9)+\frac { (0.4)(0.4-1) }{ 6 } (1)\)
⇒ y = 35 + 2.8 + (0.2)(-0.6)(9) + \(\frac { (0.4)(-0.6)(-1.6) }{ 6 } \)
⇒ y = 35 + 2.8 - 1.08 + 0.064
⇒ y_(x=1955) = 36.784
Hence, the population in the year 1955 is 36.784 lakhs.

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

Here the intervals are unequal. By Lagrange’s interpolation formula we have
x₀ = 5, x₁ = 6, x₂ = 9, x₃ = 11
y₀ = 12, y₁ = 13, y₂ = 14, y₃ = 16
\(y=f(x)=\frac { \left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 0 }-{ x }_{ 1 } \right) \left( x_{ 0 }-{ x }_{ 2 } \right) \left( { x }_{ 0 }-{ x }_{ 3 } \right) } \times { y }_{ 0 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 1 }-{ x }_{ 0 } \right) \left( x_{ 1 }-{ x }_{ 2 } \right) \left( { x }_{ 1 }-{ x }_{ 3 } \right) } \times { y }_{ 1 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 2 }-{ x }_{ 0 } \right) \left( x_{ 2 }-{ x }_{ 1 } \right) \left( { x }_{ 2 }-{ x }_{ 3 } \right) } \times { y }_{ 2 }+\frac { \left( x-{ x }_{ 0 } \right) \left( x-{ x }_{ 2 } \right) \left( x-{ x }_{ 3 } \right) }{ \left( { x }_{ 3 }-{ x }_{ 0 } \right) \left( x_{ 3 }-{ x }_{ 1 } \right) \left( { x }_{ 3 }-{ x }_{ 2 } \right) } \times { y }_{ 3 }\)
\(=\frac { (x-6)(x-9)(x-11) }{ (5-6)(5-6)(5-11) } (12)+\frac { (x-5)(x-9)(x-11) }{ (6-5)(6-9)(6-9) } (13)+\frac { (x-5)(x-6)(x-11) }{ (9-5)(9-6)(9-11) } (14)+\frac { (x-5)(x-6)(x-9) }{ (11-5)(11-6)(11-9) } (16)\)
Put x = 10
\({ y }_{ (10) }=f\left( 10 \right) =\frac { 4(1)(-1) }{ (-1)(-4)(-6) } (12)+\frac { (5)(1)(-1) }{ (1)(-3)(-5) } (13)+\frac { 5(4)(-1) }{ 4(3)(-2) } (14)+\frac { (5)(4)(1) }{ 6(5)(2) } (16)\)}
= \(\frac { 1 }{ 6 } \left( 12 \right) -\frac { 13 }{ 3 } +\frac { 5\left( 14 \right) }{ 3\times 2 } +\frac { 4\times 16 }{ 12 } \)
= 14.6663

Let the missing entries by y₁ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ⁴y_k = 0 ⇒ (E - 1)⁴y_k = 0
(E⁴ - 4E³ + 6E²- 4E + 1) y_k = 0
Put k = 0 in (1) we get,
(E₄ - 4E₃ + 6E₂ - 4E + 1)y₀ = 0
⇒ y₄ - 4y₃ + 5y₂ - 4y₁ + y₀ = 0
⇒ y₄ - 4(15) + 6(8) - 4(y₁) + 0 = 0
⇒ y₄ - 60 + 48 - 4y₁ = 0
⇒ y₄ - 4y₁ - 12 = 0
⇒ y₄ - 4y₁ = 12  (2)
put k = 1 in (1) we get
(E⁴- 4E³ + 6E² - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ + y₁ = 0
⇒ 35 - 4y₄ + 6 (15) - 4 (8) +y₁ = 0
⇒ 35 - 4y₄ + 90 - 32 +y₁ = 0
⇒ 4y₄ +y₁ = -93  (3)

⇒ y₁ = \(\frac{-45}{-15}\) = 3
⇒ y₁ = 3.
Substituting y₁ = 3 in (2) we get,
y₄ - 4(3) = 12
y₄ - 12 = 12
⇒ y₄ = 12 + 12
⇒ y₄ = 24
Hence the missing entries are 3 and 24.

By partial fraction method
\(\frac { 1 }{ (x+1)(x+2) } =\frac { A }{ x+1 } +\frac { B }{ x+2 } \)

⇒ 1 = A (x + 2) + B (x + 1)
when x = -1, 1 =  A [-1 + 2] ⇒ 1 = A
when x = -2, 1 = B [-2+ 1] ⇒1 = - B
⇒ B = -1.
∴ \(\frac { 1 }{ (x+1)(x+2) } =\frac { 1 }{ x+1 } +\frac { 1 }{ x+2 } \)
\(\therefore \triangle \left[ \frac { 1 }{ (x+1)(x+2) } \right] =\triangle \left[ \frac { 1 }{ x+1 } -\frac { 1 }{ x+2 } \right] \)
∴ \(\left( \frac { 1 }{ x+1+1 } -\frac { 1 }{ x+1 } \right) -\left( \frac { 1 }{ x+1+2 } -\frac { 1 }{ x+2 } \right) \) [∵ ∆ f(x) = f(x+1) - f(x)]
= \(\left( \frac { 1 }{ x+2 } -\frac { 1 }{ x+1 } \right) -\left( \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right) \) where h = 1
= \(\frac { 1 }{ x+2 } -\frac { 1 }{ x+1 } -\frac { 1 }{ x+3 } +\frac { 1 }{ x+2 } \)
= \(\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } -\frac { 1 }{ 1+3 } \)
= \(\frac { 2(x+1)(x+3)-1(x+2)(x+3)-1(x+1)(x+2) }{ (x+1)(x+2)(x+3) } \)
= \(\frac { 2({ x }^{ 2 }+4x+3)-({ x }^{ 2 }+5x+6)-({ x }^{ 2 }+3x-2) }{ (x+1)(x+2)(x+3) } \)
= \(\frac { -2 }{ (x+1)(x+2)(x+3) } \)
\(\therefore\ =\triangle \left[ \frac { 1 }{ (x+1)(x+2) } \right] =\frac { -2 }{ (x+1)(x+2)(x+3) } \)

Since five values are given, the polynomial which fits the data is of degree four.
Hence Δ⁵y_k = 0 (i.e) (E−1)⁵y_k = 0
i.e., (E⁵ - 5E⁴ + 10E³ - 10E² + 5E - 1)y_k = 0
E⁵y_k - 5E⁴y_k+ 10E³y_k- 10E²y_k+ 5Ey_k - y_k = 0 (1)
Put k = 0 in (1)
E⁵y₀ - 5E⁴y₀+ 10E³y₀- 10E²y₀+ 5Ey₀ - y₀ = 0
y₅ − 5y₄ + 10y₃ − 10y₂ +5 y₁ - y₀ = 0
y₅ − 5(350) +10y₃−10(260)+5(220)− 200 = 0
y₅ + 10y₃ = 3450 (2)

Put k = 1 in (1)
E⁵y₁ - 5E⁴y₁+ 10E³y₁- 10E²y₁+ 5Ey₁- y₀ = 0
y₆ − 5y₅ + 10y₄ −1 0y₃ − y₁ = 0
430  −5y₅ +10(350) −10y₃ + 5(260)− 220 = 0
5y₅+10y₃ = 5010 (3)
(3) – (2) ⇒ 4y₅ = 1560
y₅ = 390
From (1)  390 +10y₃ = 3450
10y₃ = 3450 – 390
y₃ ≅ 306

We will use Newton’s backward interpolation formula to find the polynomial.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)

To find y in terms of x
\(\therefore\) x_n + nh = x, x_n = 7, h = 1 \(\Rightarrow\) n = x − 7
\({ y }_{ (x) }=29+(x-7)(7)+\frac { (x-7)(x-6) }{ 2 } (1)\)
= \(29+7x-49+\frac { 1 }{ 2 } \left( { x }^{ 2 }-13x+42 \right) \)
= \(\frac { 1 }{ 2 } \left[ { x }^{ 2 }+x+2 \right] \)

Let age = x and premium = y
To find y at x = 63
So apply Newton’s backward interpolation formula
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)

\({ y }_{ (x=6.5) }=68.48+\frac { \frac { -2 }{ 5 } }{ 1! } \left( -6 \right) +\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) }{ 2! } 2.84+\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) \left( \frac { -2 }{ 5 } +2 \right) }{ 3! } \left( -1.16 \right) +\frac { \frac { -2 }{ 5 } \left( \frac { -2 }{ 5 } +1 \right) \left( \frac { -2 }{ 5 } +2 \right) \left( \frac { -2 }{ 5 } +3 \right) }{ 3! } (0.68)\)
= 68.48 + 2.4 – 0.3408 + 0.07424 – 0 – 0.028288
y(63) = 70.437

Since the required value is at the end of the table, apply backward interpolation formula

.\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 7.5
\(\therefore\) xn + nh = 7.5, xn = 8, h = 1 \(\Rightarrow\) n = –0.5
\({ y }_{ (x=7.5) }=512+\frac { -0.5 }{ 1! } 169+\frac { 0.5(-0.5+10 }{ 2! } 42+\frac { -0.5(-0.5+1)(-0.5+2) }{ 3! } 6\)
= 421.88

Since the pressure required is at the end of the table, we apply Backward interpolation formula. Let temperature be x and the pressure be y.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 175
\(\therefore\) x_n + nh = 175 , x_n = 180, h = 10 \(\Rightarrow\) n = −0.5

\({ y }_{ (x=1750 }=10.225+\left( -0.5 \right) (2.149)+\frac { (-0.5)(0-5) }{ 2! } (0.375)+\frac { (-0.5)(0-5)(1.5) }{ 3! } (0.049)+\frac { (-0-5)(0.5)(1.5)(2.5) }{ 4! } (0.002)\)

= 10.225−1.0745−0.046875−0.0030625 − 0.000078125
= 9.10048438
= 9.1

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
By Partial fraction method
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\frac { A }{ x+3 } +\frac { B }{ x+2 } \)
\(A=\frac { 5x+12 }{ x+12 } [x=-3]=\frac { -15+12 }{ -1 } =\frac { -3 }{ -1 } =-3\)
\(B=\frac { 5x+12 }{ x+3 } \)[x = -2] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+..\)
To find y at x = 1946
\(\therefore\) x₀ + nh = 1946, x₀ = 1941, h = 10
1941 + n(10) = 1946 \(\Rightarrow\) n = 0.5

Here we find the population for year 1946. (i.e) the value of y at x = 1946. Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\(​​{ y }_{ \left( x=1946 \right) }=20+\frac { 0.5 }{ 1! } (4)+\frac { 0.5(0.5-1) }{ 2! } (1)+\frac { 0.5(0.5-1)(0.5-2) }{ 3! } (1)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3) }{ 4! } (0)+\frac { 0.5(0.5-1)(0.5-2)(0.5-3)(0.5-4) }{ 5! } (-9)\)
= 20+2-0.125+0.0625-0.24609
= 21.69 lakhs

Let x be the weight and y be the number of students.
Difference table of cumulative frequencies are given below

Let us calculate the number of students whose weight is below 70. For this we use forward difference formula
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+....\)
To find y at x = 70
\(\therefore\) x₀+nh = 70, x₀ = 40, h = 20
40+n(20) = 70 \(\Rightarrow\) n = 1.5
\({ y }_{ \left( x=70 \right) }=250+1.5\left( 120 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) }{ 2! } \left( -20 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) }{ 3! } \left( -10 \right) +\frac { \left( 1.5 \right) \left( 0.5 \right) \left( -0.5 \right) \left( -1.5 \right) }{ 4! } \left( 20 \right) \)
= 250 + 180 - 7.5 + 0.625 + 0.46875
= 423.59
\(\cong \) 424.
Number of students whose weight is between
60 and 70 = y(70)−y(60) = 424−370 = 54

Let x be the marks and y be the number of students
By converting the given series into cumulative frequency distribution, the difference table is as follows.

\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\frac { n }{ n! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\frac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
To find y at x = 45
\(\therefore\) x₀+nh = 45 , x0 = 40, h = 10 \(\Rightarrow n=\frac { 1 }{ 2 } \)
y_{(x = 45)} = \(31+\frac { 1 }{ 2 } \times 42+\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) }{ 2 } (9)+\cfrac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) }{ 6 } \times \left( -25 \right) +\frac { \frac { 1 }{ 2 } \left( \frac { -1 }{ 2 } \right) \left( \frac { -3 }{ 2 } \right) \left( \frac { -5 }{ 2 } \right) }{ 24 } \times \left( -37 \right) \)
= \(31+21-\frac { 9 }{ 8 } -\frac { 25 }{ 16 } -\frac { 37\times 15 }{ 384 } \)
= 47.867 ≅ 48

Since we apply the forward interpolation formula,last four values of f(x) are taken into consideration (Take the values from x = 3).
The forward interpolation formula is
\({ y }_{ (x={ x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } \Delta { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{0 }+\frac { n(n-1)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 }+...\)
x₀ + nh = 3.2, x₀ = 3,y = 1
\(\therefore n=\frac { 1 }{ 5 } \)
The difference table is

y_(x=3.2) = \(16+\cfrac { 1 }{ 5 } (4)+\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) }{ 2 } \left( 0 \right) +\cfrac { \frac { 1 }{ 5 } \left( \frac { -4 }{ 5 } \right) \left( \frac { -9 }{ 5 } \right) }{ 6 } \times 10\)
= 16+0.8+0+0.48
= 17.28

To find the population for the year 1905 (i.e) the value of y at x = 1905
Since the value of y is required near the beginning of the table, we use the Newton’s forward interpolation formula.
\({ y }_{ \left( x={ x }_{ 0 }+nh \right) }={ y }_{ 0 }+\cfrac { n }{ n! } \Delta { y }_{ 0 }+\cfrac { n(n-1) }{ 2! } { \Delta }^{ 2 }{ y }_{ 0 }+\cfrac { n(n-)(n-2) }{ 3! } { \Delta }^{ 3 }{ y }_{ 0 } + ...\)
To find y at x = 1905
\(\therefore\) x₀+nh = 1905 , x₀ = 1891, h = 10
1891+n(10) = 1905 \(\Rightarrow\) n = 1.4

y_(x=1905) = \(98,752+(1.4)(33533)+\frac { (1.4)(0.4) }{ 2 } (2258)+\frac { (1.4)(0.4)(-0.6) }{ 6 } (-10435)+\frac { (1.4)(0.6)(-0.6)(-1.6) }{ 24 } (41358)\)
=  98,752 + 46946.2 + 632.4 + 584.36 + 1389.63
= 1,48,304.43
= 1,48,304