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Arithmetic Progressions | Arithmetic Progressions - Exercise 5.3

Question 30

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?

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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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Question 30

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?

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

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.

Our top 5% students will be awarded a special scholarship to Lido.

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