TRY THESE
By repeated subtraction of odd numbers starting
from 1, find whether the following numbers are
perfect squares or not? If the number is a perfect
square then find its square root.
(1) 121
(i) 55
2)36
3)49
4)90
Answers
Hey !
If the result of subtracting odd Numbers starting from 1 from a number is zero, then it is a perfect square.
Usually, sum of odd numbers is given as
S = 1 + 3 + 5 + ............+ (2n-1)
here,
first term a= 1, difference d = 2, last term l = 2n-1
l = 2n-1
=> No. of terms, n =( l+1)/2
So from AP, we get S = n²
Which means a number is perfect square only when it can be written as sum of odd numbers upto a term.
Substitute n
S = (l+1)²/4. ---------------[1]
step-1
Find What is the last odd number in the sum of odd numbers that is to be subtracted from given number.
(Means finding of l from the given sum) [✓✓✓✓ Best way]
step-2
If l value is a natural. number, then it is a perfect square.
Step-3:
To find the square root of that number, find value of n
(or)
Now,
(1) 121
Take the number as sum of odd numbers
Sum of odd numbers = 121
From S formula (1)
121 = (l+1)²/4
=> 121 ×4 = (l+1)²
=> (l+1)² = 484
=> (l+1) =√484
=> l+1 = 22
=> l = 22 - 1
=> l = 21
Here we got l value, which means 21 is a perfect square
=> 21 - (1 + 3+ 5 + 7 + .........21) = 0
Find n to get the square root of 121
i.e. , no. of terms in 1, 3, 5, 7, ...... 121
n =( l+1)/2 = 22/2 = 11.
•°• √ 121 = 11 ; 121 is a perfect square
(ii) 55
S = 55
(l+1)²/4 = 55
=> (l+1)² = 55 × 4
=> (l+1)² = 220
=> (l+1) = √220
doesn't belongs to N
•°• 55 is not a perfect square.
(iii) 36
(l+1)²/4 = 36
=> (l+1)² = 36 × 4
=> l+1 = √144
=> l+1= 12
=> l = 11
n = (11+1)/2 = 12/2 = 6
•°• 36 - (1+3+5+7+ 9+11) = 0
√36 = 6 ; 36 is perfect square
(iv) 49
(l+1)² = 49 ×4
=> (l+1)² = 196
=> l+1 = 14
=> l = 13
n = (13+1)/2 = 14/2 = 7
•°• 49 - (1+3+5+....+13) = 0
√49 = 7 ; 49 is perfect square
(v) 90
(l+1)² = 90×4
=> (l+1)² = 360
=> (l+1) = √360
90 is not a perfect square.
Meowwww xD.