$(a)\;AP,d=2\qquad(b)\;GP,r=2\qquad(c)\;AP,d=5\qquad(d)\;GP,r=3$

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Answer : (b) GP ,r=2

Explanation :

$S_{n}=\frac{2^n-1}{3}\qquad\;S_{n-1}=\frac{2^(n-1)-1}{3}$

$T_{n}=S_{n}-S_{n-1}=\frac{2^n-2^(n-1)}{3}$

$T_{n-1}=\frac{2^(n-1)-2^(n-2)}{3}$

$T_{n}-T_{n-1}=d=\frac{2^n-2*2^(n-1)+2^(n-2)}{3}=\frac{2^(n-2)}{3}$

$T_{n}-T_{n-1}=d=\frac{2^n-2*2^(n-1)+2^(n-2)}{3}=\frac{2^(n-2)}{3}$

$r=\frac{T_{n}}{T_{n-1}}=\frac{2^n-2*2^(n-1)/3}{2^(n-1)-2^(n-2)/3}=2$

Since d is a function of n but r is independent of n, it is a GP with a common ratio r=2.

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