Reema being a plant lover decides to open a nursery and she bought a few plants with pots. She wants to place pots in such a way that the number of pots in the first row is 3, in the second row is 5 and in the third row is 7 and so on.a) If Reema wants to place 120 pots in total then the number of rows formed in this arrangement is?
b) Find the difference in number of pots placed in 8th row and 3rd row.​

Given :- the number of pots in the first row is 3, in the second row is 5 and in the third row is 7 and so on.

To Find :- a) If Reema wants to place 120 pots in total then the number of rows formed in this arrangement is?

b) Find the difference in number of pots placed in 8th row and 3rd row.

Solution :-

as we can see that, 3 , 5 , 7 ________ form an AP series .

• First term = 3 .
• common difference = 5 - 3 = 2 .

So,

• Sn = (n/2)[2a + (n - 1)d]
• Sn = 120 pots .

then,

→ 120 = (n/2)[2*3 + (n-1)2]

→ 120*2 = n(6 + 2n - 2)

→ 240 = n(4 + 2n)

→ 240 = 2n(2 + n)

→ 120 = n² + 2n

→ n² + 2n - 120 = 0

→ n² + 12n - 10n - 120 = 0

→ n(n + 12) - 10(n + 12) = 0

→ (n + 12)(n - 10) = 0

→ n = (-12) or 10 .

since negative value of n is not Possible.

Therefore, The Number of rows formed is 10. (Ans. a)

Now,

→ Pots in 3rd row = T(3) = a + (3 - 1)2 = 3 + 2*2 = 3 + 4 = 7 pots .

→ Pots in 8th row = T(8) = a + (8 - 1)2 = 3 + 7*2 = 3 + 14 = 17 pots .

Therefore,

→ Required difference = 17 - 7 = 10