Question

An organization plans to plant saplings in 25 streets in a town in such a way that one sapling for the first street, two for the second, four for the third, eight for the fourth street and so on. How many saplings are needed to complete the work?

Answer 1

Hint: We have 1 sapling for 1st street, 2 sapling for the 2nd street, 4 sapling for the 3rd street and 8 sapling for the 4th street. We can write 4 as \[{{2}^{2}}\] and 8 as \[{{2}^{3}}\] . We can say that nth street is nth term of a G.P whose first term is 1 and common ratio is 2. Now, find the summation of 25 terms of this G.P.

Complete step-by-step answer:
According to the question, it is given that an organization is planning to plant saplings in 25 streets in a town in such a way that one sapling for the first street, two for the second, four for the third, eight for the fourth street. We can write 4 as \[{{2}^{2}}\] and 8 as \[{{2}^{3}}\] .
Number of sapling for the 1st street = 1
Number of sapling for the 2nd street = 2
Number of saplings for the 3rd street = 4 = \[{{2}^{2}}\] .
Number of saplings for the 3rd street = 8 = \[{{2}^{3}}\] .
We can see that we have got a G.P series whose first term is 1 and common ratio 2.
The nth term of the G.P can be written as, \[{{T}_{n}}=a.\,{{r}^{n-1}}\] ……………(1)
Now, we have to find the total number of saplings required for planting in 25 streets. It means that we have to find the summation of saplings required to plant for 1st street, 2nd street, 3rd street,………. 25th street.
In other words we can say that we have to find the summation of 25 terms of the G.P whose first term is 1 and common ratio is 2.
We know that summation of nth term of G.P is \[{{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{(1-r)}\] , where a is the 1st term and r is the common ratio.
Here, we have, a=1 and r=2.
Now, summation of 25 terms of the G.P.
\[{{S}_{25}}=\dfrac{1(1-{{2}^{25}})}{(1-2)}\]
\[{{S}_{25}}={{2}^{25}}-1\]
Hence, the total number of saplings that are required to complete the work is \[{{2}^{25}}-1\] .

Note: In this question, one can find the summation of G.P using the formula,
\[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\] . But this is wrong. This summation formula is for A.P series not G.P series.
We have to find the summation of G.P, so the formula, \[{{S}_{n}}=\dfrac{n}{2}[2a+(n-1)d]\] must not be used.