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EXERCISE 2.4 · Q67

Q.Find the nnth term and hence find the 8th term of the H.P.: 12,15,18,111,…\dfrac12, \dfrac15, \dfrac18, \dfrac1{11}, \ldots

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Reciprocals 2,5,8,11,…2,5,8,11,\ldots: A.P. with a=2,d=3a=2, d=3, so tn(A.P.)=2+(n−1)3=3n−1t_n(\text{A.P.})=2+(n-1)3=3n-1. Hence H.P. tn=13n−1t_n=\dfrac1{3n-1}. $t_8=\dfrac1{3(8)-1}=\d …

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