An arithmetic progresssion has the first term 4 and a common difference of 3. let Tn be the term n of the progression .find the smallest val

"Solution:

Tn = T1 ®^n-1 > 103

T1 = 4

r = 3

T1 = 4(3)^1-1 > 103

4^3(0) > 103

4 > 103; false

T2 = 4(3)^2-1 > 103

4^3(1) > 103

4(3) > 103

12 > 103; false

T3 = 4(3)^3-1 > 103

4^3(2) > 103

4(9) > 103

36 > 103; false

T4 = 4(3)^4-1 > 103

4^3(3) > 103

4(27) > 103

108 > 103; true

Therefore:

Answer: n = 4"

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