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  • Can someone explain Let a,b and c be the 7th,11th and 13th terms respectively of a non-constant A.P.If these are also the three consecutive terms of a G.P., then a/c is equal to :

Let a,b and c be the 7th,11th and 13th terms respectively of a non-constant A.P.If these are also the three consecutive terms of a G.P., then a/c is equal to :

  • Option 1)

     

    2

  • Option 2)

     

    7/13

  • Option 3)

     

    1/2

  • Option 4)

     

    4

Answers (1)

best_answer

 

General term of an A.P. -

T_{n}= a+\left ( n-1 \right )d

- wherein

a\rightarrow First term

n\rightarrow number of term

d\rightarrow common difference

 

 

 

Selection of terms in G.P. -

If we have to take three terms in GP, we take them as \frac{a}{r},a,ar

- wherein

Extension : If we have to take (2K+1) term in GP, we take them as

\frac{a}{r^{k}},\frac{a}{r^{k-1}},---------\frac{a}{r},a,ar------ar^{k}

From the concept 

\\a = A + 6d \\b= A + 10d \\c = A + 12d

and a, b and c are in G.P

\\\Rightarrow (A+10d)^2 = (A+6d)(A+12d)\\\Rightarrow\frac{A}{d} = -14 \\ \frac{a}{c} = \frac{A+6d}{A+12d} = \frac{6 + \frac{A}{d}}{12 + \frac{A}{d}} = \frac{6-14}{12 -14} = 4

 

 


Option 1)

 

2

Option 2)

 

7/13

Option 3)

 

1/2

Option 4)

 

4
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