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Asked by reddypogurakesh66 | 22 Jan, 2019, 11:04: AM
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Only one question will be answered per posting. I select Qn. No. 23 to answer
 
if 6 geometric means are inserted between 14 and -7/64, then the last term (-7/64) is the 8th term of geometric progression
 
if r is the common ration, then 8th term is given by,   t8 = 14×r7 = (-7/64)
 
hence r7 = -7/(64×14) = -1/128 = -(1/2)7  , hence r = -1/2
 
hence, third G.M. or 4th term, t4 = ar3 = 14×(-1/2)3 = -14/8 = -7/4
Answered by Thiyagarajan K | 22 Jan, 2019, 12:26: PM

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