**Setting up :** The instalments form an A.P.
$$a = 1000, \qquad d = 100$$

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### (i) Amount paid in the 30th instalment

**Step 1 — Use $a_n = a + (n-1)d$ with $n = 30$.**
$$a_{30} = 1000 + (30-1)(100) = 1000 + 29 \times 100$$

**Step 2 — Calculate.**
$$a_{30} = 1000 + 2900 = ₹\,3900$$

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### (ii) Amount paid in the last (40th) instalment

**Step 3 — Use $a_n = a + (n-1)d$ with $n = 40$.**
$$a_{40} = 1000 + 39 \times 100 = 1000 + 3900 = ₹\,4900$$

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### (iii) (a) Amount still to be paid after the 30th instalment

**Step 4 — Find the total amount paid in 30 instalments using $S_n = \dfrac{n}{2}\left[2a + (n-1)d\right]$.**
$$S_{30} = \frac{30}{2}\left[2(1000) + 29(100)\right] = 15\left[2000 + 2900\right]$$

**Step 5 — Calculate.**
$$S_{30} = 15 \times 4900 = ₹\,73{,}500$$

**Step 6 — Subtract from the total loan.**
$$\text{Amount remaining} = 1{,}18{,}000 - 73{,}500 = ₹\,44{,}500$$

$$\boxed{₹\,44{,}500}$$

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### (iii) (b) Ratio of the 10th instalment to the last instalment

**Step 7 — Find the 10th instalment.**
$$a_{10} = 1000 + 9 \times 100 = ₹\,1900$$

**Step 8 — Last instalment (40th) from part (ii).**
$$a_{40} = ₹\,4900$$

**Step 9 — Form the ratio and simplify.**
$$a_{10} : a_{40} = 1900 : 4900 = 19 : 49$$

$$\boxed{19 : 49}$$