Question

sum of three consecutive terms which are in ap is 27 and the sum of their square is 293 find the terms​

Answer 1

Step-by-step explanation:

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Answer 2

Answer

5, 9, 14 or 14, 9, 5

Given:

Sum of 3 consecutive A.P term = 27

Sum of their square = 293

To find:

The terms

Solution:

let, the terms be a - d, a, a + d.

[Where 'a' is the first term and 'd' is common difference.]

According to question,

$(a + d) + (a) + (a + d) = 27 \\ \\ a + d + a + a - d = 27 \\ \\ 3a = 27 \\ \\ a \: = \: 9$

again,

${(a - d)}^{2} + {a}^{2} + {(a + d)}^{2} = 273 \\ \\ {(9 - d)}^{2} + {(9)}^{2} + {(9 + {d}) }^{2} = 293 \\ \\ 81 - 18d + {d}^{2} + 81 + 81 + 18d + {d}^{2} \\ \\ 2 {d}^{2} + 243 = 293 \\ \\ 2 {d}^{2} = 293 - 243 \\ \\ 2 {d}^{2} = 50 \\ \\ {d}^{2} = 25 \\ \\ </p><p>d = ±5</p><p>$

Either, Terms are .....

9 - 5, 9 , 9+5

4, 9 , 14

[When , d = 5]

Or, Terms are ........

9+5, 9 , 9 - 5

14, 9 , 4

[When, d = -5 ]

Hence the required terms are,

5, 9, 14 or 14, 9, 5