If $x,1,z$ are in A.P. and $x,2,z$ are in G.P., then $x,4,z$ are in

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Medium

1 of 1

Application +4 | -1

x,1,z

are in A.P. and

x,2,z

are in G.P., then

x,4,z

are in

View answer and explanation

Correct option: C

H.P.

Option C (H.P.) is correct because

x, 4, z

satisfy the condition for being in Harmonic Progression, which is

4 = \frac{{2xz}}{{x + z}}

Practice All Chapter MCQs →

A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}