If the sum of two extreme numbers of an A.P. | vedclass

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(B) Let the four terms of the $A.P.$ be $(a - 3d), (a - d), (a + d), (a + 3d)$.According to the problem, the sum of the two extreme terms is $8$:$(a -

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$7$

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(B) Let the four terms of the $A.P.$ be $(a - 3d), (a - d), (a + d), (a + 3d)$.According to the problem, the sum of the two extreme terms is $8$:$(a - 3d) + (a + 3d) = 8$$2a = 8 \Rightarrow a = 4$.The product of the two middle terms is $15$:$(a - d)(a + d) = 15$$a^2 - d^2 = 15$.Substituting $a = 4$ into the equation:$4^2 - d^2 = 15$$16 - d^2 = 15$$d^2 = 1 \Rightarrow d = 1$ (taking positive value for the series).The four terms are:$a - 3d = 4 - 3(1) = 1$$a - d = 4 - 1 = 3$$a + d = 4 + 1 = 5$$a + 3d = 4 + 3(1) = 7$The terms are $1, 3, 5, 7$. The greatest number is $7$.

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

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