(C) Let the first term be $b$ and the common ratio be $r$.For an infinite $G.P.$,the sum $S$ is given by $S = \frac{b}{1 - r}$,where $|r| < 1$.Given $

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(C) Let the first term be $b$ and the common ratio be $r$.For an infinite $G.P.$,the sum $S$ is given by $S = \frac{b}{1 - r}$,where $|r| Given $S = 5$,we have $\frac{b}{1 - r} = 5$.This implies $b = 5(1 - r)$.Since $-1 If $r \to 1$,then $b \to 5(1 - 1) = 0$.If $r \to -1$,then $b \to 5(1 - (-1)) = 5(2) = 10$.Thus,for $-1 < r < 1$,the value of $b$ lies in the interval $(0, 10)$.

Class 11 Mathematics Sequences and Series nth term of special series, Sum to n terms and Infinite number of terms

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If $b$ is the first term of an infinite $G.P.$ whose sum is $5$,then $b$ lies in the interval

JEE MAIN 2018 All JEE MAIN

A
$(-\infty, -10)$
B
$(10, \infty)$
C
$(0, 10)$
D
$(-10, 0)$

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nth term of special series, Sum to n terms and Infinite number of terms

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If $b$ is the first term of an infinite $G.P.$ whose sum is $5$,then $b$ lies in the interval

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