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

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(B) Let the sequence be $a, b, c, d$.Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $b, c, d$ are in $A.P.$ with common difference $6$,we have

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$8$

Vedclass · Answer

(B) Let the sequence be $a, b, c, d$.Since $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $b, c, d$ are in $A.P.$ with common difference $6$,we have $c - b = 6$ and $d - c = 6$.Thus,$c = b + 6$ and $d = c + 6 = b + 12$.Given that the first and last terms are equal,$a = d$,so $a = b + 12$,which implies $b = a - 12$.Substituting $b = a - 12$ into $c = b + 6$,we get $c = (a - 12) + 6 = a - 6$.Now,substitute $b$ and $c$ 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 $d = a$,the last term is $8$.

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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