On January 1st, 2023, a person saved ₹ 1. On January 2nd, 2023, he saved ₹ 2 more than that on the previous day. On January 3rd, 2023, he saved ₹ 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube?

1. **Identify the Pattern**: The savings are ₹1, ₹3, ₹5... This is an Arithmetic Progression of odd numbers. As seen in PYQs, recognizing APs is key [20].\n\n2. **Apply Summation Rule**: The total savings for 'n' days is the sum of the first 'n' odd numbers. Using the AP Sum concept [36], the sum is n². So, on the 8th day, total savings is 8², on the 9th day, 9², etc.\n\n3. **Analyze the Condition**: The Total Savings (n²) must be a 'perfect square' AND a 'perfect cube'.\n - Since it is n², it is already a perfect square.\n - We just need n² to be a perfect cube.\n - This implies 'n' itself must be a perfect cube (because (a³)² = a⁶ = (a²)³).\n\n4. **Check Options**: We look for a date 'n' that is a perfect cube.\n - (A) 7th Jan: 7 is not a cube.\n - (B) 8th Jan: 8 is a perfect cube (2³). Let's verify: Total = 8² = 64. Is 64 a cube? Yes, 4³ = 64. This matches the specific PYQ knowledge about the number 64 [13][50].\n - (C) 9th Jan: 9 is a square, not a cube.\n\n Therefore, 8th January is the answer.