Five numbers are in an AP with a common diffe | vedclass

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(A) Let the five numbers in $AP$ be $(a-2d), (a-d), a, (a+d), (a+2d)$, where $d 
eq 0$.Given that the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in $

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the $5^{th}$ term is always $0$.

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(A) Let the five numbers in $AP$ be $(a-2d), (a-d), a, (a+d), (a+2d)$, where $d 
eq 0$.Given that the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in $GP$.Therefore, $(a-d)^2 = (a-2d)(a+d)$.Expanding both sides:$a^2 - 2ad + d^2 = a^2 - ad - 2d^2$.Simplifying the equation:$-2ad + d^2 = -ad - 2d^2$.$3d^2 - ad = 0$.$d(3d - a) = 0$.Since $d 
eq 0$, we have $3d - a = 0$, which implies $a = 3d$.Substituting $a = 3d$ into the terms:$(3d-2d), (3d-d), 3d, (3d+d), (3d+2d) \Rightarrow d, 2d, 3d, 4d, 5d$.Wait, let us re-evaluate the condition: $1^{st}$, $3^{rd}$, $4^{th}$ terms are in $GP$.Terms are $T_1 = a-2d, T_3 = a, T_4 = a+d$.Condition: $T_3^2 = T_1 	imes T_4$.$a^2 = (a-2d)(a+d) = a^2 + ad - 2ad - 2d^2 = a^2 - ad - 2d^2$.$0 = -ad - 2d^2 \Rightarrow ad = -2d^2$.Since $d 
eq 0$, $a = -2d$.The terms are: $(-2d-2d), (-2d-d), -2d, (-2d+d), (-2d+2d) \Rightarrow -4d, -3d, -2d, -d, 0$.Thus, the $5^{th}$ term is always $0$.

Five numbers are in an $AP$ with a common difference $d \neq 0$. If the $1^{st}$, $3^{rd}$, and $4^{th}$ terms are in a $GP$, then:

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