Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) √3, √6, √9, √12 …
(ii) 12, 32, 52, 72 …
(iii) 12, 52, 72, 73 …
(i) √3, √6, √9, √12 …
a2 – a1 = √6-√3
= √3×√2-√3 = √3(√2-1)
a3 – a2 = √9-√6 = 3-√6
= √3(√3-√2)
a4 – a3 = √12 – √9 = 2√3 – √3×√3
= √3(2-√3)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(ii) 12, 32, 52, 72 …
Or, 1, 9, 25, 49 …..
a2 − a1 = 9−1 = 8
a3 − a2 = 25−9 = 16
a4 − a3 = 49−25 = 24
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(iii) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
a2 − a1 = 25−1 = 24
a3 − a2 = 49−25 = 24
a4 − a3 = 73−49 = 24
Since, an+1 – an or the common difference is same every time.
Therefore, d = 24 and the given series forms a A.P.
Hence, next three terms are;
a5 = 73+24 = 97
a6 = 97+24 = 121
a7 = 121+24 = 145