BITSAT 2018 Slot 2 Solved Paper

Given, a,b,c are in GP.
∴b2=ac
and 2(log‌2‌b−log‌3‌c)=log‌a−log‌2‌b+log‌3‌c −log‌a
⇒b2=ac and 2b=3c
⇒c=

4a

9

and b=

2a

3

∵a+b=

5a

3

>c,b+c=

10a

9

>a
and c+a=

13a

9

>b.
∴a,b,c are the sides of a triangle.
Also, a is the greatest side
∴cos‌A=

b2+c2−a2

2bc

=−

29

48

<0
∴ΔABC is an obtuse angled triangle.
Hence, option (d) is correct.