Let the first three terms 2, p and q,with q n | vedclass

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(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$ and the $G.P.$ be $2, p, q, \dots$Given that $2, p, q$ are the $7^{	ext{th}}, 8^{	ext{th}}, 13^{	ext{th

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$163$

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(D) Let the $A.P.$ be $a, a+d, a+2d, \dots$ and the $G.P.$ be $2, p, q, \dots$Given that $2, p, q$ are the $7^{	ext{th}}, 8^{	ext{th}}, 13^{	ext{th}}$ terms of the $A.P.$:$2 = a + 6d \quad \dots(i)$$p = a + 7d \quad \dots(ii)$$q = a + 12d \quad \dots(iii)$Subtracting $(i)$ from $(ii)$,we get $p - 2 = d$.Subtracting $(ii)$ from $(iii)$,we get $q - p = 5d$.Since $q - p = 5(p - 2)$,we have $q = 6p - 10$.Since $2, p, q$ are in $G.P.$,$p^2 = 2q$.Substituting $q = 6p - 10$,we get $p^2 = 2(6p - 10) \implies p^2 - 12p + 20 = 0$.Solving for $p$,$(p - 10)(p - 2) = 0$,so $p = 10$ or $p = 2$.If $p = 2$,then $q = 2$,which contradicts $q 
eq 2$. Thus,$p = 10$.Then $d = p - 2 = 8$ and $a = 2 - 6(8) = -46$.The $G.P.$ is $2, 10, 50, 250, 1250, \dots$The $5^{	ext{th}}$ term of the $G.P.$ is $2 	imes 5^4 = 1250$.Let this be the $n^{	ext{th}}$ term of the $A.P.$: $1250 = a + (n - 1)d$.$1250 = -46 + (n - 1)8 \implies 1296 = (n - 1)8 \implies n - 1 = 162 \implies n = 163$.

Let the first three terms $2, p$ and $q$,with $q \neq 2$,of a $G.P.$ be respectively the $7^{\text{th}}$,$8^{\text{th}}$ and $13^{\text{th}}$ terms of an $A.P.$ If the $5^{\text{th}}$ term of the $G.P.$ is the $n^{\text{th}}$ term of the $A.P.$,then $n$ is equal to

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