Question

In a nursery, 37 plants have been arranged in the First row, 35 in the second, 33 in the third and so on. If there are 5 plants in the last row, how many plants are there in the nursery?
Please solve this prblm ​

Answer 1

there will be 357 plant as the no.of plants in consecutive rows are following a patter. like there are 37 in first row then 35 in second and so on. so no. of plants are decreasing by 2. so according to that it should be 37+35+33+31+29+27+25

+23+21+19+17+15+13+11

+9+7+5

=357

HOPE THIS HELPS

Answer 2

There are 357 plants are there in the nursery

Explanation:

The given system of planting plants in rows following Arithmetic Progression as first row has 37 plants , second row has 35 plants , third row has 33 plants and so on.

First term : a=37

The common difference between terms : d= -2

nth term of AP : $a(n) = a+(n-1)d$              (1)

Since the last row has 5 plants , so we need to find n for a(n) =5

Put all values in (1) , we get

$5 = 37+(n-1)(-2)$  

$5-37=-2(n-1)\\\\\Rightarrow\ -32=-2(n-1)\\\\\Rightarrow\ n-1=\dfrac{32}{2}=16\\\\\Rightarrow\ n= 16+1=17$

Sum of first n terms =$S(n)=\dfrac{n}{2}(a+l)$ , where l= last term

Total plants = $S(17)=\dfrac{17}{2}(37+5)=\dfrac{17}{2}\times42=357$

Hence, there are 357 plants are there in the nursery.

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