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Let a,b and c be in G.P. with common ratio r where a0 and 0<r1/2. If 3a, 7b and  15c are the first three terms of an A.P., then the 4th term of this A.P. is:

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a
7a/3
b
a
c
5a
d
2a/3

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detailed solution

Correct option is B

Given, a, band  c  are in G.P. with common ratio r.

 b=ar and  c=ar2  ……………… …(i)

Also, 3a,7b and 15c are in A.P.            [Given]

    3a+15c=14b 

3a+15ar2=14ar          [Using (i)] 

15r214r+3=015r29r5r+3=0(3r1)(5r3)=0

r=1/3  or 3/5r=1/3            [r(0,1/2]]

    The required A.P. is  3a,7a/3,5a/3,

Hence, 4th term is 3a/3 = a

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