एक गुणोत्तर श्रेणी के तीन पदों का गुणन फल 512 है । यदि प्रथम पद में 8 तथा द्वितीय पद में 6 जोड़ दे तो नये पदो से एक समांतर श्रेणी बनती है तीनों पदो को ज्ञात कीजिए ।
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let the term of gp be a/r, a, r
product of gp = 512
a/r×a×ar=512
a³=512
a³=8³
a=8
term of gp be 8/r, 8, 8r
8/r+8, 8+6, 8r be in ap
8/r+8, 14, 8r be in ap
so,
14-(8/r+8)= 8r-14
14-8/r-8=8r-14
6-8/r= 8r-14
8r+8/r= 20
(8r²+8)/r= 20
8r²+8=20r
8r²-20r+8=0
4(2r²-5r+2)=0
2r²-5r+2=0
2r²-4r-r+2=0
2r(r-2)-1(r-2)=0
(2r-1)(r-2)=0
r=2 r=1/2
hence term of gp be 8/r, 8, 8r (if r=2)
8/2, 8, 8×2
4, 8, 16
again term of gp be 8/r, 8, 8r if (r=1/2)
8/1/2 , 8, 8×1/2
16, 8, 4
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Answer:
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Step-by-step explanation:
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