Question

The sum of three consecutive numbers in A. P is equal to 90.if the difference of the squares of the greatest and smallest number is 480.Find the numbers

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\green{\underline \bold{Given :}}$

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• The sum of three consecutive numbers in A. P is equal to 90

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• Difference of the squares of the greatest and smallest number is 480

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$\red{\underline \bold{To \: Find:}}$

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• The numbers

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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Let the 3 consecutive terms be 'a - d' , 'a' , 'a + d' -----(1)

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Where,

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• a = First term of AP

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• d = Common difference

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$\purple{\underline \bold{According \: to \: the \ question :}}$

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Sum of three consecutive numbers in A. P is equal to 90

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So,

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$\purple\longrightarrow$ $\sf a - d + a + a + d = 90$

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$\sf \longmapsto 3a = 90$

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$\sf \longmapsto a = \dfrac { 90 } { 3 }$

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$\bf \dashrightarrow a = 30$-------(2)

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Also, Given that

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Difference of the squares of the greatest and smallest number is 480.

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Clearly greatest among the terms is 'a + d' & smallest is 'a - d'

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$\sf \longmapsto (a + d)^2 - (a - d)^2 = 480$

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$\sf \longmapsto a ^2 + d ^2 + 2ad - ( a ^2 + d ^2 - 2ad ) = 480$

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$\sf \longmapsto a ^2 + d ^2 + 2ad - a ^2 - d ^2 + 2ad ) = 480$

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$\sf \longmapsto 4ad = 480$

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$\sf \longmapsto ad = \dfrac { 480 } { 4 }$

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$\bf \longmapsto ad = 120$ ----(3)

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$\underline{\bold{\texttt{Putting a = 30 from (2) to (3)}}}$

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$\sf \longmapsto 30 \times d = 120$

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$\sf \longmapsto d = \dfrac { 120 } { 30 }$

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$\bf \dashrightarrow d = 4$

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• a = 30

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• d = 4

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$\underline{\bold{\texttt{Putting a = 30 \& d = 4 in (1)}}}$

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Numbers are '30 - 4' , '30' , '30 + 4'

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$\underline{\bold{\texttt{Consecutive numbers are :}}}$

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• 26

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• 30

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• 34

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Hence three consecutive numbers are 26 , 30 , 34

$\rule{200}5$

Answer 2

Answer:

mrpadaku is always right.....