Question

The sum of 3 consecutive term of an AP is
30° &e their product is 360°. Find the terms
elt. (d is positive)
​

Answer 1

Given: The sum of three Consecutive terms of an AP is 30 and the product of these three terms is 360.

Need to find: The three terms?

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❍ Let's Consider, the three Consecutive terms of AP are a – d, a, a + d respectively.

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$⠀⠀

• We're given that, the sum of these three terms is 30.

$:\implies\sf a - d + a + a + d = Sum \\\\\\:\implies\sf a - \cancel{\;d} + a + a + \cancel{\;d}= 30 \\\\\\:\implies\sf 3a = 30 \\\\\\:\implies\sf a = \cancel\dfrac{30}{3} \\\\\\:\implies\underline{\boxed{\pmb{\frak{a = 10}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Second\:term\:(a)\:of\:AP\:is\:{\pmb{10}}.}}}$

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• Also, second condition is that the product of these terms is 360.

$:\implies\sf \Big(a - d\Big) \Big(a~\Big)\Big(a + d\Big) = Product \\\\\\:\implies\sf a\Big(a^2 - d^2\Big) = 360 \\\\\\:\implies\sf 10\Big(10^2 - d^2\Big) = 360 \\\\\\:\implies\sf 100 - d^2 = \cancel\dfrac{360}{10} \\\\\\:\implies\sf 100 - d^2 = 36 \\\\\\:\implies\sf d^2 = 100 - 36 \\\\\\:\implies\sf d^2 = 64 \\\\\\:\implies\sf d = \sqrt{64} \\\\\\:\implies\underline{\boxed{\pmb{\frak{d = 8}}}}\;\bigstar$

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$\therefore\:{\underline{\sf{Common\:diffrence\:(d)\:of\:AP\:is\:{\pmb{8}}.}}}$⠀

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✇ Therefore, The three Consecutive terms of A.P are :

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$\;\;\;\;\dashrightarrow\sf a - d,\;a,\;a + d\\\\\\\;\;\;\;\dashrightarrow\sf 10 - 8,\;10,\;10 + 8\\\\\\\;\;\;\;\dashrightarrow\underline{\boxed{\pmb{\frak{\pink{2, \: 10, \: 18}}}}}\;\bigstar$

Answer 2

Given :-

The sum of 3 consecutive term of an AP is  30 & their product is 360

To Find :-

terms

Solution :-

Let the terms be a - d, a, a + d

Now,

$\sf a - d, a, a+d=30$

$\sf a-d+a+a+d=30$

$\sf (a+a+a) +(d-d) = 30$

$\sf 3a=30$

$\sf a=\dfrac{30}{3}$

$\sf a=10$

Now, Product is 360

$\sf a-d\times a\times a+d = 360$

$\sf a(a^2 -d^2)=360$

$\sf 10\{(10)^2 - d^2\} =360$

$\sf 10\{100 - d^2\} = 360$

$\sf 100-d^2 = \dfrac{360}{10}$

$\sf 100 - d^2 = 36$

$\sf -d^2 = 36-100$

$\sf -d^2 = -64$

$\sf d^2=64$

$\sf \sqrt{d^2} = \sqrt{64}$

$\sf d=8$

Finding terms

${\tt{\red{\underline{1^{st}\;term}}}}$

$\sf a-d$

$\sf 10-8$

$\sf 2$

${\tt{\red{\underline{2^{nd}\;term}}}}$

$\sf a$

$\sf 10$

${\tt{\red{\underline{3^{rd}\;term}}}}$

$\sf a+d$

$\sf 10+8$

$\sf 18$