Question

in a GP sum of 2nd and 4th terms is 30 the difference of 6 th and 2 nd terms is 90 find a term of a GP whose common ratio is greater than 1

Answer 1

Solution:

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Given:

$a_2 + a_4 = 30$ ..........................(i)

$a_6 - a_2 = 90$  ..........................(ii)

r > 1,.

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We know that,

=> $a_n =a_1r^{n-1}$

=> $a_2 = a(r^{2-1}) = ar$

=> $a_4 = a(r^{4-1}) = ar^3$

=> $a_2 + a_4 = 30$

=> [tex]ar + ar^3 =30 [/tex]

=> $ar(1+r^2) =30$
 
=> $1+ r^2 = \frac{30}{ar}$ ......(iii)

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(i) + (ii) => $(a_2 + a_4) +(a_6 -a_2) = 30 + 90$

=> $a_4 +a_6 =120$ ..........(iv)

=> $ar^3 +ar^5 =120$

=> $ar^3(1+ r^2) = 120$

=>$1+r^2 = \frac{120}{ar^3}$ .....(v)

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By equation equation (iii) & equation (v),.

=> $\frac{30}{ar} = \frac{120}{ar^3}$

=> $30 = \frac{120}{r^2}$

=> $30r^2 = 120$

=> $r^2 = 4$

=>$r =2$         (r≠-2, as r > 1),

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Substituting value of r, in equation (iii)
we get,

=>  $1 + r^2 = \frac{30}{ar}$

=> [tex]1+ (2)^2 = \frac{30}{a(2)} [/tex]

=> $1 + 4 = \frac{15}{a}$

=> $5 = \frac{15}{a}$

=> 5a = 15,

=> a = 3,..
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GP = 3,3(2), 3(2)², 3(2)³,

GP = 3,6,12,24,..

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                                 Hope it helps !!