If sum of two extreme numbers of an A.P. with four terms is 8 and product of remaining two middle te

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8. Sequences and Series

easy

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

A

$5$

B

$7$

C

$9$

D

$11$

Solution

(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 $7.$

Standard 11

Mathematics

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