Question

Find the first three consecutive terms in G.P. whose sum is 7 and the sum of their reciprocal is 7/4???​

Answer 1

Let the first three consecutive terms of the G. P. be a/r, a, ar.

Their sum = a/r + a + ar = 7

a ( 1/r + 1 + r ) = 7

a = 7 ÷ ( 1/r + 1 + r )

( r^2 + r + 1 ) /r = 7/a --> ( i )

Sum of their Reciprocal = r/a + 1/a + 1/ar = 7/4

( r^2 + r + 1 ) ÷ ar = 7/4

7/ ( a × a ) = 7/4

a^2 = 4

a = 2

Putting value of 'a' in Eqn ( i ),

( r^2 + r + 1 ) / r = 7/ 2

2r^2 + 2r + 2 = 7r

2r^2 - 5r + 2 = 0

2r^2 - 4r - r + 2 = 0

2r ( r - 2 ) - 1( r - 2) = 0

( 2r - 1) ( r - 2) = 0

r = 2, 1/2

So, admissible value of ' r' is 2.

First three consecutive terms ,

a/r = 2/2 = 1

a = 2

ar = 2 × 2 = 4

$\fbox {Terms\:=\:1,2,4...}$

Answer 2

Answer:

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