Sequences And Series Ques 102
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)
Show Answer
Solution:
Formula:
HARMONICAL PROGRESSION (H.P.)](/important-formula/mathematics/sequence_and_series)
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.
BETA
