Question

The sum of three consecutive terms of an A. P. is 30 and their product is 360. Find the terms.

Answer 1

let the terms be
$(a - d) \: \: \: (a) \: \: and \: (\: a + d) \\ sum = 30 \\ a - d + a + a + d = 30 \\ 3a = 30 \\ a = 10 \\ \\ product = 360 \\ (a - d)(a)(a + d) = 360 \\ a( {a}^{2} - {d}^{2} ) = 360 \\ 10(100 - d {}^{2} ) = 360 \\ 100 - d {}^{2} = 36 \\ d {}^{2} = 64 \\ d = 8 \: or \: - 8 \\ \\ numbers \: are \\ (10 - 8) \: \: (10) \: and \: (10 + 8) \\ = 2 \: \: 10 \: \: \: and \: 18 \\ \\ or \\ \\ (10 - ( - 8)) \: \: (10) \: and \: (10 + ( - 8)) \\ \\ = 18 \: \: 10 \: \: and \: 2$

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Answer 2

Answer:

Three consecutive terms of an A. P are : $18 , 10 , 2$  or  $2, 10 , 18$ .

Step-by-step explanation:

To find : three consecutive terms of an A. P.

Given : the sum of three consecutive terms of an A. P. is $30$ and their product is $360$ .

Solution :

• As per given data we know that , sum of three consecutive terms of an A. P. is $30$ and their product is $360$ .  
• Let , $a-d ,a , a + d$ be the three terms of A.P .
• We are given that , sum of three consecutive terms is $30$ that is :

         $a -d + a +a + d= 30 \\\\3a = 30 \\\\ a = 10$

• We are also given that , product of terms is $360$ , that is :

        $( a-d )(a)(a + d )= 360\\\\a(a^{2}- d^{2}) = 360 \\\\10 (10^{2}- d^{2} ) = 360 \\ \\(100 - d^{2})= \frac{360}{10}\\\\100 -d^{2} =36 \\\\100- 36 = d^{2} \\\\64 = d^{2}$

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

         $d =$ ± $8$

• Now , if , $a = 10$ and $d = -8$ , then three consecutive terms of an A. P are :

        $a - d = 10 -(-8)= 18 \\\\a = 10 \\\\a + d = 10 + (-8) = 2$

• If , $a = 10$ and $d = 8$ , then three consecutive terms of an A. P are :  

         $a - d = 10 -(8)= 2 \\\\a = 10 \\\\a + d = 10 + (8) = 18$