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NCERT Solutions Class 10 Mathematics Solutions for Arithmetic Progressions - Exercise 5.3 in Chapter 5 - Arithmetic Progressions

Question 29 Arithmetic Progressions - Exercise 5.3

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. How many logs are in the top row?

Answer:

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…

For the given A.P.,

First term, a = 20 and common difference, _d_ = _a_2−_a_1 = 19−20 = −1

Let a total of 200 logs be placed in _n_ rows.

Thus, Sn = 200

By the sum of nth term formula,

\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\ 200=\frac{\mathrm{n}}{2}(2(20)+(\mathrm{n}-1)(-1)) \\ 400=n(40-n+1) \\ 400=n(41-n) \\ 400=41 n-n 2 \\ n^{2}-41 n+400=0 \\ n^{2}-16 n-25 n+400=0 \\ n(n-16)-25(n-16)=0 \\ (n-16)(n-25)=0 \\ \text { Either }(n-16)=0 \text { or } n-25=0 \\ n=16 \text { or } n=25

By the nth term formula,

an = a+(n−1)d

a16 = 20+(16−1)(−1)

a16 = 20−15

a16 = 5

Similarly, the 25th term could be written as;

a25 = 20+(25−1)(−1)

a25 = 20−24

a25 = –4

It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.

Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

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