17. 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, 5 in second row and
7 in third row and so on.
(a) How many pots are placed in 10th row.
(i) 19 (ii) 20 (iii) 21 (iv) none of these
(b) Find the difference in number of pots placed in 3rd row and 8th row?
(i) 12 (ii) 10 (iii) 15 (iv) 16
(c) If she has enough space for 16 rows, how many pots are placed in 16th
row
(i) 30 (ii) 31 (iii) 32 (iv) 33
(d) Find the sum in number of pots placed in 2nd row and 6th row
(i) 19 (ii) 20 (iii) 17 (iv) none of these
(e) If for an AP an = 4n + 5 find the 5th term.
(i) 15 (ii) 25 (iii) 20 (iv) none of these
Answers
Solution :
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, 5 in second row and 7 in third row and so on.
We can clearly observe that the series, 3, 5, 7 and so on forms an Arithmetic Progression ( or AP ) series .
Here , the starting number of pots , corresponding to the value of the starting term a is 3 .
The common difference d becomes a_n - a_(n-1)
=> 5 - 3 = 7 - 5 = 2 .
Hence , d = 2 .
For the 10th row
a_10 = a + 9d
=> 3 + 18
=> 21 .
Hence, Option (iii) is correct .
The difference between the number of pots present in the 3rd and 8th row -
=> a_8 - a_3
=> ( a + 7d ) - ( a + 2d )
=> 5d
=> 5 × 2
=> 10 pots.
Hence , Option (ii) is correct .
Number of pots present in the 16th row
=> a + 15d
=>3 + 30
=> 33 pots.
Hence, Option (iv) is correct .
Sum of number of pots present in the 2nd and 6th row
=> a_2 + a_6
=> ( a + d ) + ( a + 5d )
=> 2 ( a + 3d )
=> 2 ( 9 )
=> 18 pots .
Hence, Option (iv) is correct .
a_n = 4n + 5
n = 5 ;
a_5 = 4 × 5 + 5
=> 20 + 5
=> 25 .
Hence , Option (ii) is correct .
This is the required answer .
_____________________________________________