If the first term of a Geometric Progression | vedclass

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(B) Let the first term be $a = 1$ and the common ratio be $r$.The $n^{th}$ term of a $GP$ is given by $a_n = a \cdot r^{n-1}$.The third term is $a_3 =

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$\pm 3$

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(B) Let the first term be $a = 1$ and the common ratio be $r$.The $n^{th}$ term of a $GP$ is given by $a_n = a \cdot r^{n-1}$.The third term is $a_3 = 1 \cdot r^{3-1} = r^2$.The fifth term is $a_5 = 1 \cdot r^{5-1} = r^4$.Given that $a_3 + a_5 = 90$,we have $r^2 + r^4 = 90$.Rearranging the equation: $r^4 + r^2 - 90 = 0$.Let $x = r^2$,then $x^2 + x - 90 = 0$.Factoring the quadratic equation: $(x + 10)(x - 9) = 0$.This gives $x = -10$ or $x = 9$.Since $x = r^2$,$r^2 = -10$ (not possible for real $r$) or $r^2 = 9$.Therefore,$r = \pm 3$.

If the first term of a Geometric Progression $(GP)$ is $1$ and the sum of its third and fifth terms is $90$,find the common ratio.

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