Question

The sum of three consecutive numbers in ap is equal to 90 if the difference of the square 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

Okay. This was a bit tricky (at least to me) and so I decided to do it by taking different values of n and finding a pattern in the answers if it exists.

Case n=1:

Numbers = 1,2,3

No. of ways to select 3 numbers = 3C3 = 1

Selections which are AP = 1 (1,2,3)

Case n=2:

Numbers = 1,2,3,4,5

No. of ways to select 3 numbers = 5C3=10

Selections which are AP = 4 (1,2,3 ; 2,3,4 ; 3,4,5 ; 1,3,5)

Case n=3:

Numbers = 1,2,3,4,5,6,7

No. of ways to select 3 numbers = 7C3 = 35

Selections which are AP = 9 (1,2,3 ; 2,3,4 ; 3,4,5 ; 4,5,6 ; 5,6,7 ; 1,3,5 ; 2,4,6 ; 3,5,7 ; 1,4,7)

Similarly, I found that

for n=4 => Selections which are AP = 16 (Square of 4)

for n=5 => Selections which are AP = 25 (Square of 5)

And so I drew the following pattern from these results.

Let us say you have x=2n+1 consecutive numbers and you are talking about an AP of k (here k=3) numbers.

Let d be the common difference in AP

Maximum no. of AP you can get with d=1 => (x+1)-k.

For example, when 2n+1=7(n=3) and k=3, AP with d=1 are (1,2,3) (2,3,4) (3,4,5) (4,5,6) (5,6,7), i.e. 5.

(7+1)-3=5.

Maximum no. of AP you can get with d=2 => (x+1)-(k+(k-1))

For example, when 2n+1=7 (n=3) and k=3, AP with d=2 are (1,3,5) (2,4,6) (5,6,7), i.e. 3.

(7+1)-(3+2)=3

Maximum no. of AP you can get with d=3 => (x+1)-(k+2(k-1))

For example, when 2n+1=7 (n=3) and k=3, AP with d=3 is (1,4,7), i.e. 1.

(7+1)-(3+4)=1

And so on.

Thus we observe that for any n when k=3, the no. of such AP are sum of n odd numbers.

n=1 => 1

n=2 => 1+3=4

n=3 => 1+3+5=9

n=4 => 1+3+5+7=16

n=5 => 1+3+5+7+9=25

and so on.

Therefore,

Total number of ways to select 3 numbers out of 2n+1 numbers = (2n+1)C3

Selections which are AP = n square.

=> Probability = n*n/(2n+1)C3.

Thank you. :)