Four numbers are in arithmetic progression. T | vedclass

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Question

(d) Let ${A_1},{A_2},{A_3}$ and ${A_4}$ are four numbers in $A.P.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(d) Let ${A_1},{A_2},{A_3}$ and ${A_4}$ are four numbers in $A.P.$

${A_1} + {A_4} = 8$ ..$(i)$

and ${A_2}.\,{A_3} = 15$ ..$(ii)$

The sum of terms equidistant from the beginning and end is constant and is equal to

sum of first and last terms.

Hence, ${A_2} + {A_3} = {A_1} + {A_4} = 8$ ..$(iii)$

From $(ii)$ and $(iii),$

${A_2} + \frac{{15}}{{{A_2}}} = 8$

==> $A_2^2 - 8{A_2} + 15 = 0$

${A_2} = 3\,\,{m{or}}\,\,5$ and ${A_3} = 5\,\,\,{m{or}}\,\,{m{3}}$.

As we know, ${A_2} = \frac{{{A_1} + {A_3}}}{2}$

==> ${A_1} = 2{A_2} - {A_3}$

==> ${A_1} = 2 	imes 3 - 5 = 1$ and ${A_4} = 8 - {A_1} = 7$

Hence the series is, $1, 3, 5, 7.$

So that least number of series is $1.$

Four numbers are in arithmetic progression. The sum of first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is

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