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  • Match the questions given under Column I with their appropriate answers given under the Column II.a)4,1,1/4,1/16 i)A.P

Match the questions given under Column I with their appropriate answers given under the Column II.

 

Column-I Column-II
a)4,1,\frac{1}{4}.\frac{1}{16} i) A.P.
b)2,3,5,7 ii) Sequence
c)13,8,3,-2,-7 iii)G.P.

 

Answers (1)

\left( a \right) \text{ Given 4, 1, }\frac{1}{4},\frac{1}{16}\text{~~ Here,}\frac{a_{2}}{a_{1}}=\frac{1}{4}~,\frac{a_{3}}{a_{2}}=\frac{\frac{1}{4}}{1}=\frac{1}{4}\text{~ and similarly}\frac{a_{4}}{a_{3}}=\frac{1}{16} *\frac{4}{1}=\frac{1}{4}~~ \\\\

   Hence, the given numbers is in G.P with common ratio  \frac{1}{4}~~ \\\\


   \\~ \left( a \right) - \left( iii \right) ~~~~ \\\\ \left( b \right) \text{~Given~2, 3, 5, 7 } \\\\ ~Here,~a_{2}-a_{1}=3-2=1~~~ a_{3}-a_{2}=5-3=2~~ a_{4}-a_{3}=7-5=2~~ \\\\
  ~a_{2}-a_{1} \neq a_{3}-a_{2}~~ \\\\

    Hence, it is not in AP

 

  Now, we will check the ratio \frac{a_{2}}{a_{1}}=\frac{3}{2},~\frac{a_{3}}{a_{2}}=\frac{5}{7}~~ \\\\


 \\ ~So,\frac{a_{2}}{a_{1}} \neq \frac{a_{3}}{a_{2}}\text{~~Thus,~it~is not a G.P } \\\\ ~ \left( b \right) - \left( ii \right) ~~ \\\\ ~ \left( c \right) ~13,8,~3, -2, -7 Here, a_{2}-a_{1}=8-13=-5 \\\\


  \\~a_{3}-a_{2}=3-8=~-5 \\\\ ~a_{4}-a_{3}=-2-3=-5~ \\\\ ~~~~a_{2}-a_{1}=a_{3}-a_{2}\text{ Hence, it is an A.P } \\\\ ~~~ \left( c \right) - \left( i \right) \\\\

 

 

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