Question

14. Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their
squares is 180.

Answer 1

Hey there !!


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

▶Now,

A/Q,

=> ( a - d ) + a + ( a + d ) = 18.

=> a - d + a + a + d = 18.

=> 3a = 18.

=> a = $\frac{18}{3} .$

=> a = 6.

And,

=> ( a - d )² + a² + ( a + d )² = 180.

=> ( a² + d² - 2ad ) + a² + ( a² + d² + 2ad ) = 180.

=> a² + d² - 2ad + a² + a² + d² + 2ad = 180.

=> 3a² + 2d² = 180.

=> 3 × 6² + 2d² = 180. [→ a = 6 ].

=> 3 × 36 + 2d² = 180.

=> 108 + 2d² = 180.

=> 2d² = 180 - 108.

=> 2d² = 72.

=> d² = $\frac{72}{2} .$

=> d² = 36.

=> d = √36.

=> d = ±6.


▶ When taking, a = 6 and d = 6, we get

→ a - d = 6 - 6 = 0.

→ a = 6.

→ a + d = 6 + 6 = 12.


And, when taking, a = 6 and d = -6, we get

→ a - d = 6 - (-6) = 6 + 6 = 12.

→ a = 6.

→ a + d = 6 + (-6) = 6 - 6 = 0.


✔✔ Hence, the required number are ( 0,6,12 ) or ( 12,6,0 ). ✅✅

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Answer 2

$\huge{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$

$\bf{see \: in \: pic}$


$\huge \boxed{ \ulcorner{hope \: it \: helps}}$


$\huge{ \mathbb{THANKS}}$