Question

Find five number in Arithmetic progressions whose sum is 25 and the sum of whose square is 135
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Answer 1

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

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

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the numbers be a-2d, a-d, a, a+d, a+2d

A/q

$\mapsto\sf{a-2d+a-d+a+a+d+a+2d=25}\\\\\mapsto\sf{5a \cancel{-3d+3d}=25}\\\\\mapsto\sf{5a=25}\\\\\mapsto\sf{a=\cancel{\dfrac{25}{5} }}\\\\\mapsto\sf{\pink{a=5}}$

So;

Putting the value of (a) in this number :

$\mapsto\sf{(a-2d)^{2} +(a-d)^{2} +a^{2} +(a+d)^{2} +(a+2d)^{2} =135}\\\\\mapsto\sf{a^{2} +4d^{2} \cancel{-4ad}+a^{2} +d^{2} \cancel{-2ad}+a^{2} +a^{2} +d^{2} \cancel{+2ad}+a^{2} +4d^{2} \cancel{+4ad}=135}\\\\\mapsto\sf{5a^{2} +10d^{2} =135}\\\\\mapsto\sf{5(5)^{2} +10d^{2}=135}\\ \\\mapsto\sf{5\times 25+10d^{2} =135}\\\\\mapsto\sf{125+10d^{2} =135}\\\\\mapsto\sf{10d^{2} =135-125}\\\\\mapsto\sf{10d^{2} =10}\\\\\mapsto\sf{d^{2} =\cancel{\dfrac{10}{10} }}\\\\\mapsto\sf{d^{2} =1}$

$\mapsto\sf{\pink{d=\pm1}}$

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

Answer 2

Answer:-

Let the numbers be a - 2d , a - d , a , a + d , a + 2d.

given,

Sum of the numbers = 25.

So,

a - 2d + a - d + a + a + d + a + 2d = 25

5a = 25

a = 25/5

a = 5

And also given that,

Sum of the squares of the numbers = 135

(a - 2d)² + (a - d)² + (a)² + (a + d)² + (a + 2d)² = 135

(5 - 2d)² + (5 - d)² + (5)² + (5 + d)² + (5 + 2d)² = 135

(5)² - 2(5)(2d) + (2d)² + (5)² - 2(5)(d) + (d)² + 25 +(5)²

+ 2(5)(d) + d² + (5)² + 2(5)(2d) + (2d)² = 135

25 - 20d + 4d² + 25 - 10d + d² + 25 + 25 + 10d + d² + 25 + 20d + 4d² = 135

10d² + 125 = 135

10d² = 135 - 125

10d² = 10

d² = 10/10

d² = 1

d = ± 1

Therefore,

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

a - d = 5 - 1 = 4

a = 5

a + d = 5 + 1 = 6

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

Hence, the numbers are 3,4,5,6,7.