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NCERT Exemplar · Q2

Q.A man saved Rs 6600066000 in 2020 years. In each succeeding year after the first year he saved Rs 200200 more than what he saved in the previous year. How much did he save in the first year?

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The problem is an arithmetic series where the total saved over 20 years is Rs 66,000, with each year’s saving increasing by Rs 200. Using the sum formula for an arithmetic progression, the first year’s saving is found to be Rs 1,400.

The key here is recognising that the yearly savings form an arithmetic progression (AP). In an AP, the difference between consecutive terms is constant — in this case, Rs 200. The total saved over 20 years is the sum of the first 20 terms of this AP. We are asked for the first term, which is the saving in the first year.

Let’s denote the saving in the first year as aa (in rupees). Then:

  • Year 1: aa
  • Year 2: a+200a + 200
  • Year 3: a+400a + 400
  • …
  • Year 20: a+19×200=a+3800a + 19 \times 200 = a + 3800

The sum of the first nn terms of an AP is given by:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]

where nn is the number of terms, aa is the first term, and dd is the common difference.

Here, n=20n = 20, d=200d = 200, and S20=66000S_{20} = 66000.

  1. Write the sum formula with the given values:

66000=202[2a+(20−1)×200]66000 = \frac{20}{2} \left[ 2a + (20-1) \times 200 \right]

  1. Simplify the factor outside: 202=10\frac{20}{2} = 10, so:

66000=10[2a+19×200]66000 = 10 \left[ 2a + 19 \times 200 \right]

  1. Compute 19×20019 \times 200: 19×200=380019 \times 200 = 3800, so:

66000=10[2a+3800]66000 = 10 \left[ 2a + 3800 \right]

  1. Divide both sides by 10:

6600=2a+38006600 = 2a + 3800

  1. Isolate 2a2a:

2a=6600−3800=28002a = 6600 - 3800 = 2800

  1. Solve for aa:

a=28002=1400a = \frac{2800}{2} = 1400

Watch out

A common mistake is to forget that the increase applies from the second year onward, so the 20th year’s saving is a+19da + 19d, not a+20da + 20d. Always check: for nn terms, the last term is a+(n−1)da + (n-1)d.

Tip

You can also solve by first finding the average saving per year: total divided by 20 gives Rs 3,300. In an AP, the average of all terms equals the average of the first and last terms. So a+(a+3800)2=3300\frac{a + (a+3800)}{2} = 3300, which gives a=1400a = 1400 — a quicker mental check.

✓Final answer

He saved Rs 1,400 in the first year.

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