A) 2
B) 3
C) \[\frac{3}{2}\]
D) -1
Correct Answer: A
Solution :
[a] Let the progression be a, \[a+d,a+2d,\] Then \[{{x}_{4}}=3{{x}_{1}}\Rightarrow a+3d=3a\Rightarrow 3d=2a\] ... (i) Again \[{{x}_{7}}=2{{x}_{3}}+1\] \[\Rightarrow \,\,a+6d=2(a+2d)+1\Rightarrow 2d=a+1\] ... (ii) Solving (i) and (ii) we get \[a=3,\,\,d=2\]
You need to login to perform this action.
You will be redirected in
3 sec
