Sequences and Series: Real-World Scenarios (2)

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Question

A person deposits ₹10001000 on the first day of the month. Each subsequent day, the deposit increases by ₹100100. How much will be deposited in total over 3030 days?

Solution — Step by Step

The deposits form an arithmetic progression: 1000,1100,1200,1000, 1100, 1200, \ldots with first term a=1000a = 1000 and common difference d=100d = 100.

a30=a+(n1)d=1000+29×100=3900a_{30} = a + (n-1)d = 1000 + 29 \times 100 = 3900 Sn=n2(a+an)=302(1000+3900)=15×4900=73500S_n = \frac{n}{2}(a + a_n) = \frac{30}{2}(1000 + 3900) = 15 \times 4900 = 73500

Total deposit over the month is ₹73,50073{,}500.

Final answer: ₹73,50073{,}500.

Why This Works

Whenever a quantity grows by a fixed amount each step, AP applies. The sum formula Sn=n2(first+last)S_n = \tfrac{n}{2}(\text{first} + \text{last}) is essentially Gauss’s pairing trick: pair the first and last, second and second-last, etc. Each pair sums to the same value, and there are n/2n/2 such pairs.

If the growth were multiplicative (e.g., interest compounding), it’d be a GP — different formula, same logic.

Alternative Method

Use Sn=n2[2a+(n1)d]=302[2000+2900]=15×4900=73500S_n = \tfrac{n}{2}[2a + (n-1)d] = \tfrac{30}{2}[2000 + 2900] = 15 \times 4900 = 73500. Same answer; useful when only aa and dd are given.

For “nth-term-by-formula” CBSE questions, write the AP as a,a+d,a+2d,a, a+d, a+2d, \ldots and explicitly identify aa and dd first. Skipping this step costs marks even when the final answer is right.

Common Mistake

Computing a30a_{30} as a+30da + 30d instead of a+29da + 29d. The nnth term has (n1)(n-1) steps from the first term, not nn. This off-by-one error is the most common AP slip in CBSE Class 11.

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