20. Reema being a plant lover decides to open a nursery and she bought few plants with
pots. She wants to place pots in such a way that number of pots in first row is 3, in
second row is 5& in third row is 7 and so on ......
a) If Reema wants to place 120 pots in total then the total number of rows formed in this arrangement
i) 12 ii) 10 iii) 14 iv) 8
b) According to the above question number of pots placed in last row are ?
i) 22 ii) 21 iii) 24 iv) 18 c) Find the deference in number of pots placed in 8th row & 3rd row
i) 10 ii) 11 iii) 14
d) What are the number of pots placed in 7th row?
i) 13 ii) 15 iii) 17
e) If in an A.P. an = 5+4n then the common difference
i) 5 ii) 4 iii) 9
Section – III (6× 2 = 12M) Very short answer type question :
iv) 15
iv) 21
iv) can’t be determined
Answers
Answer:
here is ur Answer
Explanation:
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 (b ans.)
Learn more :-
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