Population 2021 and 2026 by three methods

Census interval 10 years; last census 2011. For 2021, $n = 1$ decade; for 2026, $n = 1.5$ decades.

Population in 2021 (n = 1)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2021} = P_n + n\bar x = 83{,}000 + 1\times 13{,}000.0 = 96{,}000$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2021} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1} = 112{,}678$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

$P_{2021} = 99{,}348$

Population in 2026 (n = 1.5)

Arithmetical increase method

Average increase per decade: $\bar x = \dfrac{83{,}000-18{,}000}{5} = 13{,}000.0$

$$P_{2026} = P_n + n\bar x = 83{,}000 + 1.5\times 13{,}000.0 = 102{,}500$$

Geometrical increase method

Average growth rate per decade (geometric mean): $r = \left(\dfrac{83{,}000}{18{,}000}\right)^{1/5}-1 = 35.757\%$

$$P_{2026} = P_n(1+r)^n = 83{,}000\times(1+0.35757)^{1.5} = 131{,}286$$

Decreasing rate of growth method

Average decrease in rate $= 6.061\%$ per decade.

Population at 2026 is obtained by linear interpolation between the decade values: $P_{2026} = 106{,}122$.

Summary

Answer: 2021: arithmetical 96,000, geometrical 112,678, decreased rate 99,348. 2026: arithmetical 102,500, geometrical 131,286, decreased rate 106,122.