Solve;
Asked by Anish
11th October 2018,
10:36 PM
Answered by Expert
Answer:
| B | R | Y | B | R | Y | B | R | Y | B | R | Y |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
now for the first blue circle radius is 1 so area Π
second blue circle is a hollow circle so area is Π(42 -32)= 7Π
like wise 3rd blue circle area = Π(72 -62)= 13Π
so
1Π, 7Π, 13Π,......are in AP d = 6
similarly you can prove for other circles
Answered by Expert
26th October 2018,
10:58 AM
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