If the sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle term is 15, then greatest number of the series will be
Options:
A .  
5
B .  
7
C .  
9
D .  
11
Answer: Option B : B Let four numbers are a−3d,a−d,a+d,a+3d. Now (a−3d)+(a+3d)=8⇒a=4 and (a−d)(a+d)=15⇒a2−d2=15⇒d=1 Thus required numbers are 1,3,5,7. Hence greatest number is 7.
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