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

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Question

(b) Let four numbers are $a - 3d,\,a - d,\;a + d,\;a + 3d$.

Now $(a-3d)(a+3d)=8 $

$\Rightarrow $$a = 4$

and $(a - d)(a + d) = 15$

$ \Right

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(b) Let four numbers are $a - 3d,\,a - d,\;a + d,\;a + 3d$.

Now $(a-3d)(a+3d)=8 $

$\Rightarrow $$a = 4$

and $(a - d)(a + d) = 15$

$ \Rightarrow $${a^2} - {d^2} = 15$

$ \Rightarrow $$d = 1$

Thus required numbers are $1, 3, 5, 7.$

Hence greatest number is $7.$

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

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