Given a sequence of 4 numbers,the first three | vedclass

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(B) Let the sequence be $a, b, c, d$.Since the first three terms $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since the last three terms $b, c, d$ are i

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$8$

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(B) Let the sequence be $a, b, c, d$.Since the first three terms $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since the last three terms $b, c, d$ are in $A.P.$ with a common difference of $6$,we have $c - b = 6$ and $d - c = 6$.Given that the first and last terms are equal,$a = d$.From $d - c = 6$,we have $d = c + 6$. Since $a = d$,then $a = c + 6$,which implies $c = a - 6$.From $c - b = 6$,we have $b = c - 6 = (a - 6) - 6 = a - 12$.Substituting $b = a - 12$ and $c = a - 6$ into the $G.P.$ condition $b^2 = ac$:$(a - 12)^2 = a(a - 6)$$a^2 - 24a + 144 = a^2 - 6a$$144 = 18a$$a = 8$.Since $a = d$,the last term $d = 8$.

Given a sequence of $4$ numbers,the first three of which are in $G.P.$ and the last three are in $A.P.$ with a common difference of $6$. If the first and last terms in this sequence are equal,then the last term is:

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