Q1. [4] § 5.3 nth Term of an AP § 5.4 Sum of First n Terms of an AP
Your elder brother wants to buy a car and plans to take a loan from a bank for his car. He repays his total loan of ₹ 1,18,000 by paying every month, starting with the first instalment of ₹ 1,000 and he increases the instalment by ₹ 100 every month.
Based on the information given above, answer the following questions :
- (i) Find the amount paid by him in the $30^{th}$ instalment. [1]
- (ii) If the total number of instalments is 40, what is the amount paid in the last instalment ? [1]
- (iii) What amount does he still have to pay after the $30^{th}$ instalment ? [2]
Previously asked in CBSE board exam
2026 30/2/1 Q36
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding stimulus
Model Answer
The instalments form an AP where a = ₹1000, d = ₹100.
(i) 30th instalment:
$a_{30} = a + 29d = 1000 + 29 \times 100 = 1000 + 2900 = ₹3900$
(ii) Last (40th) instalment:
$a_{40} = 1000 + 39 \times 100 = 1000 + 3900 = ₹4900$
(iii) Amount paid after 30 instalments:
$S_{30} = \frac{30}{2}[2 \times 1000 + 29 \times 100] = 15 \times [2000 + 2900] = 15 \times 4900 = ₹73,500$
Amount still to be paid $= 1,18,000 - 73,500 = ₹44,500$
Source: Arithmetic Progressions, Application of AP (Sum of n terms)
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Explanation
- Key formula used: $a_n = a + (n-1)d$ for the nth term, and $S_n = \frac{n}{2}[2a + (n-1)d]$ for sum.
- For part (iii), examiners expect you to find $S_{30}$ first, then subtract from the total loan — both steps carry marks.
- Write the AP identification (a, d) clearly at the start to show your method.
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