Question

Find five number in Arithmetic progressions whose sum is 25 and the sum of whose square is 135​

Answer 1

Answer:

Let the numbers lee

a-2d,a-d,a,a+d,a+2d.

⇒a−2d,+a−d+a+a+a+a+2d=25

⇒5a=25⇒a=5

[(a−2d)2+(a−d)2+a2(a+d)2+(a+2d)2]=135

⇒(5−2d)2+(5d)2+(5+d)2(5+2d)2=135

[(a+b)2+(a−b)2=2(a2+b2)]

⇒2(25+4d)2+2(25+d)2=110

⇒25+4d2+25+d2=55

⇒5d2=5

⇒d2=1   d=±1

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Answer 2

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}} </p><p>$

Five number in Arithmetic progression whose sum is 25 and the sum of whose square is 135.

The number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}} </p><p>$

Now;

Putting the value of d = 1

a-2d = 5 - 2(1) = 5 - 2 = 3

a-d = 5 - 1 = 4

a = 5

a+d = 5 + 1 = 6

a + 2d = 5 + 2(1) = 5+2 = 7

Putting the value of d = -1

a - 2d = 5 - 2(-1) = 5 + 2 = 7

a - d = 5 - (-1) = 5 + 1 = 6

a = 5

a + d = 5 + (-1) = 5 - 1 = 4

a + 2d = 5 + 2(-1) = 5 - 2 = 3