Question

the sum of three consecutive numbers in an A.P is 21 and their product is 231.find the numbers.class X

Answer 1

so
this is the
a p
that is 3 7 11

Answer 2

$\bold{Answer:}$

$\bold{3, 7, 11}$

$\bold{Step} - \bold{by} - \bold{step\ explanation:}$

$Let the three numbers be \ \ x - d,\ x,\ x + d \\ \\ Sum \\ \\ \frac{3}{2}[x - d + x + d] = 21 \\ \\ \frac{3}{2} \times 2x = 21 \\ \\ 3x = 21 \\ \\ x = 21 \div 3 = \bold{7}$

$So we found the middle term among the three. \\ \\ Now let's find the common difference.$

$Product \\ \\ x(x - d)(x + d) = 231 \\ \\ 7(7 - d)(7 + d) = 231 \\ \\ (7 - d)(7 + d) = 231 \div 7 \\ \\ 7^2 - d^2 = 33 \ \ \ \ \ \ \ \ \ \ \ \ [(a - b)(a + b) = a^2 - b^2] \\ \\ 49 - d^2 = 33 \\ \\ d^2 = 49 - 33 \\ \\ d^2 = 16 \\ \\ d = \sqrt{16} \\ \\ d = \pm 4 \\ \\ d = 4 \ \ \ \ \ OR \ \ \ \ \ d = -4$

$If \ \ d = 4, \\ \\ x - d = 7 - 4 = \bold{3} \\ \\ x + d = 7 + 4 = \bold{11} \\ \\ \\ If \ \ d = -4, \\ \\ x - d = 7 - (-4) = 7 + 4 = \bold{11} \\ \\ x + d = 7 + (-4) = 7 - 4 = \bold{3} \\ \\ \\ \therefore\ The terms are \ \ \bold{3, 7}\ and \ \bold{11}.$

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