Answer
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Hint: We can find the total number of students in school by adding the students from class first to fifth. For this geometric progression will get used.
The general formula of geometric progression is $a,a{r^2},a{r^3}$and so on. Where$'a'$is the first term and $'r'$is the common ratio.
${n^{th}}$Term of geometric progression${a_n} = a{r^{n - 1}}$
Common ratio${r_n} = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$
Complete step by step solution:
Number of trees planted by students in standard I\[ = 2\]
Number of trees planted by students in standard II$ = 4$
Number of trees planted by students in standard III$ = 8$
Here the series of geometric progression is $2,4,8$..........where ${a_1},{a_2},{a_3}$ the elements are. Here in question five standards are given so we have to find the number of trees planted by students in standard IV and V. For this we will first find a common ratio.
Finding common ratio:${r_1} = \dfrac{4}{2} = 2$$({r_1} = \dfrac{{{a_2}}}{{{a_1}}})$
${r_2} = \dfrac{8}{4} = 2$ $({r_1} = \dfrac{{{a_3}}}{{{a_2}}})$
So, the common ratio is 2.
Finding elements ${a_4},{a_5}$
${r_3} = \dfrac{{{a_4}}}{{{a_3}}}$
By cross multiplying we will get
${a_4} = {r_3} \times {a_3}\\$
${a_4} = 2 \times 8 = 16$
${r_4} = \dfrac{{{a_5}}}{{{a_4}}}$
By cross multiplying we will get
${a_5} = {r_4} \times {a_4}\\$
${a_5} = 2 \times 16 = 32$
By this we get all the elements of the series $2,4,8,16,32$
Total number of trees planted by students in the school $ = 2 + 4 + 8 + 16 + 32 = 62$
Note:
Alternative solution: - We can simply solve this question without formula by simply observing the series and how is the pattern is succeeding for example in this question series is 2, 4, 8 and in each step factor of 2 is multiplying so fourth term will be 8 multiplied by 2 i.e., 16 and after that 16 multiplied by 2 i.e. 32 will be the last term. Also be cautious in determining whether it is a geometric progression or arithmetic progression. In arithmetic progression there will be common difference instead of common ratio.
The general formula of geometric progression is $a,a{r^2},a{r^3}$and so on. Where$'a'$is the first term and $'r'$is the common ratio.
${n^{th}}$Term of geometric progression${a_n} = a{r^{n - 1}}$
Common ratio${r_n} = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$
Complete step by step solution:
Number of trees planted by students in standard I\[ = 2\]
Number of trees planted by students in standard II$ = 4$
Number of trees planted by students in standard III$ = 8$
Here the series of geometric progression is $2,4,8$..........where ${a_1},{a_2},{a_3}$ the elements are. Here in question five standards are given so we have to find the number of trees planted by students in standard IV and V. For this we will first find a common ratio.
Finding common ratio:${r_1} = \dfrac{4}{2} = 2$$({r_1} = \dfrac{{{a_2}}}{{{a_1}}})$
${r_2} = \dfrac{8}{4} = 2$ $({r_1} = \dfrac{{{a_3}}}{{{a_2}}})$
So, the common ratio is 2.
Finding elements ${a_4},{a_5}$
${r_3} = \dfrac{{{a_4}}}{{{a_3}}}$
By cross multiplying we will get
${a_4} = {r_3} \times {a_3}\\$
${a_4} = 2 \times 8 = 16$
${r_4} = \dfrac{{{a_5}}}{{{a_4}}}$
By cross multiplying we will get
${a_5} = {r_4} \times {a_4}\\$
${a_5} = 2 \times 16 = 32$
By this we get all the elements of the series $2,4,8,16,32$
Total number of trees planted by students in the school $ = 2 + 4 + 8 + 16 + 32 = 62$
Note:
Alternative solution: - We can simply solve this question without formula by simply observing the series and how is the pattern is succeeding for example in this question series is 2, 4, 8 and in each step factor of 2 is multiplying so fourth term will be 8 multiplied by 2 i.e., 16 and after that 16 multiplied by 2 i.e. 32 will be the last term. Also be cautious in determining whether it is a geometric progression or arithmetic progression. In arithmetic progression there will be common difference instead of common ratio.
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