Answer :

(i) Put n = 1

A1= 3 + 4(1) = 7


Put n = 2


A2= 3 + 4(2) = 11


Put n = 3


A3= 3 + 4(3) = 15


Common difference, d1= a2 – a1 = 11 – 7 = 4


Common difference, d2= a3 – a2 = 15 – 11 = 4


Since, d1 = d2


Therefore, it’s an A.P. with sequence 7, 11, 15,…


(ii) Put n = 1


A1= 5 + 2(1) = 7


Put n = 2


A2= 5 + 2(2) = 9


Put n = 3


A3= 5 + 2(3) = 11


Common difference, d1= a2 – a1= 9 – 7 = 2


Common difference, d2= a3 – a2= 11 – 9 = 2


Since, d1 = d2


Therefore, it’s an A.P. with sequence 7, 9, 11,…


(iii) Put n = 1


A1= 6 – 1 = 5


Put n = 2


A2= 6 – 2 = 4


Put n = 3


A3= 6 – 3 = 3


Common difference, d1= a2 – a1= 4 – 5 = -1


Common difference, d2 = a3- a2= 3 – 4 = -1


Since, d1=d2


Therefore, it’s an A.P. with sequence 5, 4 , 3,…


(iv) Put n = 1


A1= 9 – 5(1) = 4


Put n = 2


A2= 9 – 5(2) = -1


Put n = 3


A3= 9 – 5(3) = -6


Common difference, d1= a2 – a1 = -1 – 4 = -5


Common difference,d2= a3 – a2 = -6 – (-1) = -5


Since, d1=d2


Therefore, it’s an A.P. with sequence 4, -1, -6,…


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