kailash is drawing a square pattern. he draws 1 square in the first row . he draws 2 squares in second row and 3 in third row . how many rows can be make if he draws 45 squares in total. please answer with procedure
Answers
Answered by
1
9 rows can be made in total.
In 1 row - 1 square
2 -2
3-3
4-4
5-5
6-6
7-7
8-8
9-9
Plus them al-
1+2+3+4+5+6+7+8+9= 45
In 1 row - 1 square
2 -2
3-3
4-4
5-5
6-6
7-7
8-8
9-9
Plus them al-
1+2+3+4+5+6+7+8+9= 45
lovishgoel94:
procedure will be this only??
Answered by
1
using formula of sum of n th term
s(total sum) = n/2(2a + (n - 1)d )
where s = sum or here total squares = 45
a = 1st term or here no of square in 1st row = 1
n = no of terms or here no. of rows = n(let)
d = common difference or here no of squares which increase in each row = 1
so
45 = n/2(2 + n - 1 )
⇒90= n(n + 1)
⇒n² + n - 90 = 0
doing middle term
⇒ n² +10n - 9n - 90 = 0
⇒n(n + 10) - 9(n + 10) = 0
⇒(n- 9)(n + 10 ) = 0
so n - 9 = 0⇒n = 9 and
n + 10 = 0 ⇒ n = - 10 which is not possible
so 9 rows is the answer.
s(total sum) = n/2(2a + (n - 1)d )
where s = sum or here total squares = 45
a = 1st term or here no of square in 1st row = 1
n = no of terms or here no. of rows = n(let)
d = common difference or here no of squares which increase in each row = 1
so
45 = n/2(2 + n - 1 )
⇒90= n(n + 1)
⇒n² + n - 90 = 0
doing middle term
⇒ n² +10n - 9n - 90 = 0
⇒n(n + 10) - 9(n + 10) = 0
⇒(n- 9)(n + 10 ) = 0
so n - 9 = 0⇒n = 9 and
n + 10 = 0 ⇒ n = - 10 which is not possible
so 9 rows is the answer.
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