Sequences and Series 5 Question 14
5. Suppose four distinct positive numbers $a _1, a _2, a _3, a _4$ are in GP. Let $b _1=a _1, b _2=b _1+a _2, b _3=b _2+a _3$ and $b _4=b _3+a _4$.
Statement I The numbers $b _1, b _2, b _3, b _4$ are neither in AP nor in GP.
Statement II The numbers $b _1, b _2, b _3, b _4$ are in HP.
(2008, 3M)
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Answer:
Correct Answer: 5. (c)
Solution:
- Let $a _1=1, a _2=2, \Rightarrow a _3=4, a _4=8$
$\therefore \quad b _1=1, b _2=3, b _3=7, b _4=15$
Clearly, $b _1, b _2, b _3, b _4$ are not in HP.
Hence, Statement II is false.
Statement I is already true.
