Question

the sum of three consecutive terms of an ap is 21 and the sum of the squares of this term is 165 find these terms.

Answer 1

$\textbf{Answer:-}$

Step by step explanation

Refer to the 3 attachment




$\textbf{Hope it helps!}$

Answer 2

Hey there !!

Answer:

→ 4 , 7 , 10 or 10 , 7 , 4 .

Step-by-step explanation:

Let the required numbers be ( a - d ) , a and ( a + d ) .

Then, A/Q

⇒ ( a- d ) + a + (a + d ) = 21 .

⇒ a - d + a + a + d = 21 .

⇒ 3a = 21 .

⇒ a = 21/3 .

∴ a = 7 .

And,

⇒ ( a - d )² + a² + ( a + d )² = 165 .

⇒ a² + d² - 2ad + a² + a² + d² + 2ad = 165 .

⇒ 3a² + 2d² = 165 .

⇒ 3(7)² + 2d² = 165 .   [ ∵ a = 7 ] .

⇒ 3 × 49 + 2d² = 165 .

⇒ 147 + 2d² = 165 .

⇒ 2d² = 165 - 147 .

⇒ d² = 18/2 .

⇒ d² = 9 .

⇒ d = √9 .

∴ d = ± 3 .

Therefore, the required number are :-

[ Taking d = +3 . ]

→ ( a - d ) = 7 - 3 = 4 .

→ a = 7 .

→ (a + d ) = 7 + 3 = 10 .

And, taking d = -3 .

→ ( a - d ) = 7 - ( -3 ) = 10 .

→ a = 7 .

→ (a + d ) = 7 +( -3 ) = 4 .

Hence, the required numbers are ( 4 , 7 , 10 ) or ( 10 , 7 , 4 ) .

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