Question

4. The sum of the three consecutive integers is 30.
Find the sum of their squares.​

Answer 1

★ Answer -

• The sum of the squares of the three consecutive integers is 302.

★ To find -

• The sum of the squares of the three consecutive integers whose sum is 30.

★ Step-by-step explanation -

• Before finding the sum of the squares of the three consecutive integers whose sum is 30, let's find the integers!

Let -

• The first integer be "x".

Then -

• The next two consecutive integers will be "x + 1" and "x + 2".

Given that -

• The sum of these three consecutive integers is 30.

Therefore -

$\sf\longmapsto(x) + (x + 1) + (x + 2) = 30$

$\longmapsto \sf x + x + 1 + x + 2 = 30$

$\longmapsto\sf3x + 3 = 30$

$\longmapsto \sf3x = 30 - 3$

$\longmapsto \sf3x = 27$

$\longmapsto\sf x = \cancel\dfrac{27}{3}$

$\longmapsto\sf x = 9$

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Hence, the three consecutive integers are -

$\mapsto\bf x = 9.$

$\mapsto\bf x + 1 = 9 + 1 = 10.$

$\mapsto \bf x + 2 = 9 + 2 = 11.$

Now -

• The squares of these consecutive integers are (9)², (10)² and (11)².

And the sum of their squares is -

$\tt\hookrightarrow{(9)}^{2} + {(10)}^{2} + {(11)}^{2}$

$\tt \hookrightarrow81 + 100 + 121$

$\hookrightarrow \tt302$

Thus -

• The sum of the squares of the three consecutive integers is 302.

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Answer 2

Answer:

Given :-

• The sum of three consecutive integers is 30.

To Find :-

• What is the sum of their squares.

Solution :-

Let,

$\leadsto \bf First\: Number =\: a$

$\leadsto \bf Second\: Number =\: a + 1$

$\leadsto \bf Third\: Number =\: a + 2$

According to the question,

$\bigstar$ The sum of three consecutive integers is 30.

$\footnotesize \implies \bf \bigg\{1^{st}\: Number\bigg\} + \bigg\{2^{nd}\: Number\bigg\} + \bigg\{3^{rd}\: Number\bigg\} =\: 30$

$\implies \sf a + a + 1 + a + 2 =\: 30$

$\implies \sf a + a + a + 1 + 2 =\: 30$

$\implies \sf 3a + 3 =\: 30$

$\implies \sf 3a =\: 30 - 3$

$\implies \sf 3a =\: 27$

$\implies \sf a =\: \dfrac{\cancel{27}}{\cancel{3}}$

$\implies \sf a =\: \dfrac{9}{1}$

$\implies \sf\bold{\purple{a =\: 9}}$

Hence, the required numbers are :-

✮ First Number :

➸ First Number = a

➸ First Number = 9

✮ Second Number :

➸ Second Number = a + 1

➸ Second Number = 9 + 1

➸ Second Number = 10

✮ Third Number :

➸ Third Number = a + 2

➸ Third Number = 9 + 2

➸ Third Number = 11

Now, we have to find the sum of their squares :

$\footnotesize \mapsto \bf \bigg(1^{st}\: Number\bigg)^2 + \bigg(2^{nd}\: Number\bigg)^2 + \bigg(3^{rd}\: Number\bigg)^2$

$\mapsto \sf (9)^2 + (10)^2 + (11)^2$

$\mapsto \sf (9 \times 9) + (10 \times 10) + (11 \times 11)$

$\mapsto \sf 81 + 100 + 121$

$\mapsto \sf 181 + 121$

$\mapsto \sf\bold{\red{302}}$

$\therefore$ The sum of their squares is 302.