Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) a, 2a, 3a, 4a …
(ii) a, a2, a3, a4 …
(iii) √2, √8, √18, √32 …
(i) a, 2a, 3a, 4a …
a2 – a1 = 2a–a = a
a3 – a2 = 3a-2a = a
a4 – a3 = 4a-3a = a
Since, an+1 – an or the common difference is same every time.
Therefore, d = a and the given series forms a A.P.
Hence, next three terms are;
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(ii) a, a2, a3, a4 …
a2 – a1 = a2–a = a(a-1)
a3 – a2 = a3 – a2 = a2(a-1)
a4 – a3 = a4 – a3 = a3(a-1)
Since, an+1 – an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
(iii) √2, √8, √18, √32 …
a2 – a1 = √8-√2
= 2√2-√2 = √2
a3 – a2 = √18-√8
= 3√2-2√2 = √2
a4 – a3 = 4√2-3√2
= √2
Since, an+1 – an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = √32+√2 = 4√2+√2
= 5√2 = √50
a6 = 5√2+√2
= 6√2 = √72
a7 = 6√2+√2
= 7√2 = √98