Sequence And Series Question 436
Question: Three numbers form a G.P. If the $ 3^{rd} $ term is decreased by 64, then the three numbers thus obtained will constitute an A.P. If the second term of this A.P. is decreased by 8, a G.P. will be formed again, then the numbers will be
Options:
A) 4, 20, 36
B) 4, 12, 36
C) 4, 20, 100
D) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
$ a,\ ar,\ ar^{2} $ are in G.P. $ a,\ ar-8,\ ar^{2}-64 $ are in A.P., we get
$ \Rightarrow $ $ a(r^{2}-2r+1)=64 $ …..(i) Again, $ a,\ ar-8,\ ar^{2}-64 $ are in G.P.
$ \therefore $ $ {{(ar-8)}^{2}}=a(ar^{2}-64) $ or $ a(16r-64)=64 $ …..(ii) Solving (i) and (ii), we get $ r=5,\ a=4 $ . Thus required numbers are 4, 20, 100. Trick: Check by alternates according to conditions (a)
$ \Rightarrow $ 4, 20, - 28 which are not in A.P. (b)
$ \Rightarrow $ 4, 12, - 28 which are also not in A.P. (c)
$ \Rightarrow $ 4, 20, 36 which are obviously in A.P. with 16 as common difference. These numbers also satisfy the second condition $ i.e. $ 4, 20 - 8, 36 are in G.P.
BETA
