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# Observe the following pattern and supply the missing numbers.

11^{2} = 121

101^{2} = 10201

10101^{2} = 102030201

1010101^{2} = ?

?^{2} = 10203040504030201

**Solution:**

The square of the given number has the same number of zeros before and after digit 2 as it has in the original number.

11^{2} = 121

101^{2} = 10201

10101^{2} = 1002001

1010101^{2} = 1020304030201

101010101^{2} = 10203040504030201

**☛ Check: **Class 8 Maths NCERT Solutions Chapter 6

**Video Solution:**

## Observe the following pattern and supply the missing numbers. 11^{2} = 121 101^{2} = 10201 10101^{2} = 102030201 1010101^{2} = ? ?^{2} = 10203040504030201

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 Question 5

**Summary:**

For the following pattern, 11^{2} = 121 101^{2} = 10201 1001^{2} = 1002001 100001^{2} = 1....2....1 10000001^{2} = ..........., the missing numbers are 1020304030201 and 101010101

**☛ Related Questions:**

- Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 = 132 42 + 52 + _2 = 212 52 + -2 + 302 = 312 62 + 72 + _2 = _2
- Without adding, find the sum. (i) 1 + 3 + 5 + 7 + 9 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
- (i) Express 49 as the sum of 7 odd numbers. (ii) Express 121 as the sum of 11 odd numbers.
- How many numbers lie between squares of the following numbers? (i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100

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