If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :
Solution
Let first term = a > 0
<br><br>Common ratio = r > 0
<br><br>ar + ar<sup>2</sup>
+ ar<sup>3</sup>
= 3 ....(i)
<br><br>ar<sup>5</sup>
+ ar<sup>6</sup>
+ ar<sup>7</sup>
= 243 ....(ii)
<br><br>$\Rightarrow$ r<sup>4</sup>(ar + ar<sup>2</sup> + ar<sup>3</sup>) = 243
<br><br>$\Rightarrow$ r<sup>4</sup>(3) = 243
<br><br>$\Rightarrow$ r = 3 as r > 0
<br><br>from (i)
<br><br>3a + 9a + 27a = 3
<br><br>$\Rightarrow$ a = ${1 \over {13}}$
<br><br>$\therefore$ S<sub>50</sub> = ${{a\left( {{r^{50}} - 1} \right)} \over {\left( {r - 1} \right)}}$
<br><br>= ${1 \over {26}}\left( {{3^{50}} - 1} \right)$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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