Consider the A.P a1,a2,a3…an, the G.P b1,b2…bn such that a1=b1=1:a9=b9 and ∑i=19 ar=369 then

a1=b1=1,a9=1+8d=bn=1.r8∑r=19 ar=921+a9=921+r8=369 sum of n terms A.P=n2a+l ⇒r=3b7=b1r6=1.(3)6=27

Consider the A.P a1,a2,a3…an, the G.P b1,b2…bn such that a1=b1=1:a9=b9 and ∑i=19 ar=369 then

a1=b1=1,a9=1+8d=bn=1.r8∑r=19 ar=921+a9=921+r8=369 sum of n terms A.P=n2a+l ⇒r=3b7=b1r6=1.(3)6=27

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$Consider the A.P a_{1}, a_{2}, a_{3} \dots a_{n}, the G.P b_{1}, b_{2} \dots b_{n} such that a_{1} = b_{1} = 1 : a_{9} = b_{9}$ $and \sum_{i = 1}^{9} a^{r} = 369 then$

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$b_{9} = 18$

$b_{7} = 27$

$b_{6} = 27$

$b_{8} = 81$

answer is B.

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$a_{1} = b_{1} = 1, a_{9} = 1 + 8 d = b_{n} = 1 . r^{8}$

$\sum_{r = 1}^{9} a^{r} = \frac{9}{2} (1 + a_{9}) = \frac{9}{2} (1 + r^{8}) = 369$ $(s u m o f n t e r m s A . P = \frac{n}{2} (a + l))$

$\begin{array}{l} \Rightarrow r = \sqrt{3} \\ b_{7} = b_{1} r^{6} = 1 . (\sqrt{3})^{6} = 27 \end{array}$

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Consider the A.P a1,a2,a3…an, the G.P b1,b2…bn such that a1=b1=1:a9=b9 and ∑i=19 ar=369 then

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$Consider the A.P a_{1}, a_{2}, a_{3} \dots a_{n}, the G.P b_{1}, b_{2} \dots b_{n} such that a_{1} = b_{1} = 1 : a_{9} = b_{9}$ $and \sum_{i = 1}^{9} a^{r} = 369 then$