Number of toothpicks to built the top trapezoid = 5

Number of toothpicks to built 2^nd row trapezoid = 9

Number of toothpaste to built 3^rd row trapezoid = 13

Here, we have to find how many trapezoids will be built by using 1000 toothpicks.

(i) \(\because\) 5, 9, 13,...... are A.P.

whose first term is a = 5

common difference id d = a₂ - a₁ = a₃ - a₃ = 4

Here, we have to find for which value of n.

the sum of A.P. is equal to 1000 or just less than or greater than 1000.

Sum of A.P. = \(\frac{n}{2}[2a + (n-1)d]\)

\(=\frac{n}{2}[10 + (n-1)7]\)      (\(\because\) a = 5, d = 4)

\(=\frac{n}{2}\times2[5 + 2n- 2]\)

= n (3 + 2n)

\(\because\) At n = 21, sum of A.P. 

= 21 (3 + 42)

= 21 x 45 

= 545 < 1000

But at n = 22, sum of A.P. 

= 22 (3 + 44)

= 22 x 47

= 1034 > 1000

So, we can built 21 trapezoids in last complete row with the help of 945 toothpicks and there will be 55 toothpicks still left.

To built 22 trapezoids we will need 34 more toothpick

(ii) \(\because\) a₁ = 5, a₂ = 9, a₃ = 13,.....

\(\therefore \) a_n = a + (n-1)d

\(\therefore \) a₂₁ = 5 + (21 - 1)4       (a = a₁ = 5, d = 4, n = 21)

= 5 + 20 x 4

= 85

Hence, in last complete row (21^throw) we have to use 85 toothpicks to construct trapezoids in the last complete row.