Question

1
2. In a geometric progression, the sum of
6th, 7th and 8th terms-is 17. The sum of
11th, 12th, 13th terms is 4131. Then the
sum of 7th, gth and 9th terms is ...​

Answer 1

Answer:

answer for the given problem is given

Answer 2

$Let \: 'a' \:and \: 'r' \: are \: first \:term \:and$

$Common \:ratio \: of \: a \: geometric$

$progression .$

$\boxed { \pink { n^{th} \: term ( a_{n}) = a r^{n-1} }}$

/* According to the problem given */

$i) Sum \:of \: a_{6} , a_{7} \:and \: a_{8} = 17$

$\implies a_{6} + a_{7} + a_{8} = 17$

$\implies ar^{5} + ar^{6} + ar^{7} = 17$

$\implies ar^{5}( 1 + r + r^{2} ) = 17 \: ---(1)$

$ii) Sum \:of \: a_{11} , a_{12} \:and \: a_{13} = 4131$

$\implies a_{11} + a_{12} + a_{13} = 4131$

$\implies ar^{10} + ar^{11} + ar^{12} = 4131$

$\implies ar^{10}( 1 + r + r^{2} ) = 4131\: ---(2)$

$Do \: equation \:(2) \div equation \:(1) , we \:get$

$\implies r^{5} = 243$

$\implies r^{5} = 3^{5}$

$\implies r = 3 \:--(3)$

$Now, Sum \:of \: a_{7} , a_{8} \:and \: a_{9}$

$= a_{7} + a_{8} +a_{9}$

$= ar^{6} + ar^{7} + ar^{8}$

$= ar^{6}( 1 + r + r^{2} )$

$= ar^{5}( 1 + r + r^{2} )\times r$

$= 17 \times 3 \: \blue{ [From \::(1) \:and \: (3) ]}$

$= 51$

Therefore.,

$\red{Sum \:of \: a_{7} , a_{8} \:and \: a_{9}\:terms }$

$\green { = 51 }$

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