Question

{"given":"Third term of a GP: $t_3 = ar^2 = 4$, where $a$ is the first term and $r$ is the common ratio.","key_observation":"For a GP with first term $a$ and common ratio $r$, the product of the first five terms equals $(ar^2)^5$ because the exponents of $r$ sum to $0+1+2+3+4=10$ and there are five factors of $a$, giving $a^5 r^{10} = (ar^2)^5$.","option_analysis":[{"label":"(A)","text":"$4^3$","verdict":"incorrect","explanation":"The product of the first five terms is $a^5 r^{10} = (ar^2)^5 = 4^5$, not $4^3$. $4^3$ would correspond to the product of only three symmetric terms."},{"label":"(B)","text":"$4^5$","verdict":"correct","explanation":"The five terms are $a, ar, ar^2, ar^3, ar^4$. Their product is $a^5 r^{10} = (ar^2)^5 = 4^5$."},{"label":"(C)","text":"$4^4$","verdict":"incorrect","explanation":"The product of five terms in a GP gives $(ar^2)^5 = 4^5$, not $4^4$. $4^4$ does not correspond to any natural grouping here."},{"label":"(D)","text":"None of these","verdict":"incorrect","explanation":"The answer $4^5$ is among the given options, so 'None of these' is incorrect."}],"answer":"(B)","formula_steps":[]}