Question

the sum of three number in ap is 12 and sum of their squares are 66. find the number?​

Answer 1

$\sf\large\green{\underline { }}to \: find : -$

• AP.

$\sf\large\green{\underline { }}given : -$

• sum of three number in AP is 12.
• sum of their square is 66.

$\sf\large\green{\underline { }}solution : -$

★Let us consider ,

• 1st term of AP be 'a-d'.
• Difference be 'd'.

★The three terms will be ,

• (a - d) , (a) , (a + d)

★As sum of the three numbers in AP is 12.

$\sf\large\green{\underline { }}a - d + a + a + d = 12 \\\sf\large\green{\underline { }}3a = 12 \\ \sf\large\green{\underline { a = 4}}$

★As sum kf their square is 66.

$\sf\green{\underline { }}( {a - d)}^{2} + {a}^{2} + ( {a + d)}^{2} = 66 \\ \sf\green{\underline { }} {a}^{2} - 2ad + {d}^{2} + {a}^{2} + {a}^{2} + 2ad + {d}^{2} = 66 \\ \sf\green{\underline { }}3 {a}^{2} + 2 {d}^{2} = 66 \\ \sf\green{\underline { }}3( {4)}^{2} + 2 {d}^{2} = 66 \\ \sf\green{\underline { }}3(16) + 2 {d}^{2} = 66 \\ \sf\green{\underline { }}48 + 2 {d}^{2} = 66 \\ \sf\green{\underline { }}2 {d}^{2} = 18 \\ \sf\green{\underline { }} {d}^{2} = 9 \\ \sf\green{\underline { d = 3}}$

★Therefore , a= 4 and d = 3

★Terms are....

• a-d = 4-3 = 1
• a = 4
• a + d = 4+3 = 7

AP is 1,4,7..............

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