The sum of the first two terms of a G.P. is 3 | vedclass

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(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$Given: $a_1 + a_2 = 36 \Rightarrow a + ar = 36$$\Rightarrow a(1 + r) = 36$ $...

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$\frac{3280}{81}$

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(B) Let $a$ be the first term and $r$ be the common ratio of the $G.P.$Given: $a_1 + a_2 = 36 \Rightarrow a + ar = 36$$\Rightarrow a(1 + r) = 36$ $...(1)$Also,$a_1 \cdot a_3 = 9 \cdot a_2 \Rightarrow a \cdot ar^2 = 9 \cdot ar$$\Rightarrow ar^2 = 9$ $...(2)$From $(1)$,$a = \frac{36}{1+r}$. Substituting this into $(2)$:$\frac{36}{1+r} \cdot r^2 = 9 \Rightarrow 4r^2 = 1 + r \Rightarrow 4r^2 - r - 1 = 0$Solving for $r$: $r = \frac{1 \pm \sqrt{1 - 4(4)(-1)}}{8} = \frac{1 \pm \sqrt{17}}{8}$.Wait,re-evaluating the condition: $a_1 a_3 = 9 a_2 \Rightarrow a(ar^2) = 9(ar) \Rightarrow a^2 r^2 = 9ar$. Since $a, r 
eq 0$,$ar = 9$.Substitute $ar = 9$ into $a + ar = 36$: $a + 9 = 36 \Rightarrow a = 27$.Then $27r = 9 \Rightarrow r = \frac{1}{3}$.The sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r}$.$S_8 = \frac{27(1 - (1/3)^8)}{1 - 1/3} = \frac{27(1 - 1/6561)}{2/3} = \frac{81}{2} \cdot \frac{6560}{6561} = \frac{3280}{81}$.

The sum of the first two terms of a $G.P.$ is $36$ and the product of the first and the third terms is $9$ times the second term,then find the sum of the first $8$ terms.

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