Question

R. 4 The units digit's of the 99th number in the sequence : 1, 3, 9, 27, ... is :
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Answer 1

Answer:

Note that the series can be written as

30, 31, 32, 33....... =

34n, 34n + 1, 34n + 2,, 34n + 3........ for n = 0, 1, 2, 3 ..........

And the last digit has the repeating pattern

1, 3, 9, 7 ...........

So....the 99th term is 398 = 34(24) + 2 which will end in 9

Answer 2

Answer:

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Step-by-step explanation:

Lets check whether it is AP or not,

d1 = 3 - 1 = 2

d2 = 9 - 3 = 6

Therefore it is not AP

Lets check whether it is GP or not,

r1 = 3/1 = 3

r2 = 9/3 = 3

As r1 = r2,

Therefore it is GP

tn term of GP = ar^(n-1) = 1*3^98

Therefore the unit digit is 9