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# Which of the following are harmonic progressions?(i) 1,14,17,110,...... (ii) 1,23,12,25,........ (iii) 12,16,118,......(iv) 13,17,111,........ (v) 6, 4, 3, ......... (vi) 1,12,14,......

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Solution

## In harmonic progression, the numerator of each term is set to be 1 and the set of terms at the denominator forms an A.P.So, we need to check whether the terms at denominators form an A.P.(i) 1,14,17,110,...... The terms at denominator are 1,4,7,10,...Now, 4−1=3,7−4=3,10−7=3So, they have a common difference and that is 3.∴ The denominators form an A.P.Thus, the given sequence is a H.P. (ii) 1,23,12,25,........ We can write the given sequence as 11,132,12,152,........ The terms at denominator are 1,32,2,52,...Now, 32−1=12,2−32=12,52−2=12So, they have a common difference and that is 12.∴ The denominators form an A.P.Thus, the given sequence is H.P. (iii) 12,16,118,......The terms at denominator are 2,6,18,...Now, 6−2=4,18−6=12So, there difference is not common.∴ The denominators does not form an A.P.Thus, the given sequence is not an H.P.(iv) 13,17,111,........ The terms at denominator are 3,7,11,...Now, 7−3=4,11−7=4So, they have a common difference and that is 4.∴ The denominators form an A.P.Thus, the given sequence is a H.P(v) 6, 4, 3, ......... We can write the given sequence as 116,114,113,........ The terms at denominator are 16,14,13...Now, 14−16=112,13−14=112So, they have a common difference and that is 112.∴ The denominators form an A.P.Thus, the given sequence is H.P. (vi) 1,12,14,......The terms at denominator are 1,2,4,...Now, 2−1=1,4−2=2So, there difference is not common.∴ The denominators does not form an A.P.Thus, the given sequence is not an H.P.Hence, (i), (ii), (iv) and (v) are harmonic progressions.

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