find the value of k for which the terms (2k+1) , 8 and 3k are in A.P.
Answers
Answered by
2
a=2k +1. d=8-2k+1. d1=3k-8
=9-2k. =
d=d1
9-2k=3k-8
9+8=3k+2k
17=5k
k=17/5
=9-2k. =
d=d1
9-2k=3k-8
9+8=3k+2k
17=5k
k=17/5
swarajbarwal123:
wrong
Answered by
4
take its common difference as D1 D2 D3.......
first time be A1=(2k+1)
A2 = 8
A3 =3k
here is given that it is in AP
so D1=D2
D1=A2 - A1
=8-(2K+1)
=8-2K-1
=7-2K................1
D2=A3 - A2
=3K - 8............2
HERE,
D1=D2
SO,
1=2
7-2K=3K-8
7+8=3K+2K
15=5K
K=3
first time be A1=(2k+1)
A2 = 8
A3 =3k
here is given that it is in AP
so D1=D2
D1=A2 - A1
=8-(2K+1)
=8-2K-1
=7-2K................1
D2=A3 - A2
=3K - 8............2
HERE,
D1=D2
SO,
1=2
7-2K=3K-8
7+8=3K+2K
15=5K
K=3
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