Hari deposited Rs.1000, Rs.2,000, Rs.3,000 in bank on his son Aashish's first, second and third birthday respectively. In this way, he increases the deposit by Rs. 1,000 on every birthday. a) Define…

  "The series is: 1,000, 2,000, 3,000, ..., 10,000.",
  "The **sum of an arithmetic series** is given by:",
  \[
  S_n = \frac{n}{2} \times [2a + (n-1)d].
  \]
  "Substituting the values:",
  \[
  S_{10} = \frac{10}{2} \times [2(1000) + (10-1)(1000)] = 5 \times [2000 + 9000] = 5 \times 11000 = 55,000.
  \]
  "**Total money deposited up to 10th birthday = Rs. 55,000.**"
]

}, "part_c": { "question": "c) Is Rs. 66,000 deposited on Aashish's account by 11th birthday? Give reasons.", "answer": [ "To check if Rs. 66,000 is deposited by the 11th birthday, calculate the sum up to 11 terms ((n = 11)):", [ S_{11} = \frac{11}{2} \times [2(1000) + (11-1)(1000)] = \frac{11}{2} \times [2000 + 10000] = \frac{11}{2} \times 12000 = 11 \times 6000 = 66,000. ] "Since the sum up to 11 terms is exactly Rs. 66,000, Rs. 66,000 is deposited by the 11th birthday." ] }, "part_d": { "question": "d) In how many years will Rs. 1,05,000 be deposited in the account of Aashish? Find it.", "answer": [ "We need to find the number of years ((n)) such that the sum (S_n = 105,000).", "Using the sum formula:", [ S_n = \frac{n}{2} \times [2a + (n-1)d] = 105,000, ] "Substitute (a = 1000) and (d = 1000):", [ \frac{n}{2} \times [2000 + (n-1)1000] = 105,000, ] "Simplify:", [ \frac{n}{2} \times (1000n + 1000) = 105,000, ] [ n(1000n + 1000) = 210,000, ] [ 1000n^2 + 1000n - 210,000 = 0, ] "Divide by 1000:", [ n^2 + n - 210 = 0. ] "Solve the quadratic equation using the quadratic formula (n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = 1), and (c = -210):", [ n = \frac{-1 \pm \sqrt{1 + 840}}{2} = \frac{-1 \pm \sqrt{841}}{2} = \frac{-1 \pm 29}{2}. ] "Discard the negative solution:", [ n = \frac{28}{2} = 14. ] "Rs. 1,05,000 will be deposited in 14 years." ] } }