The sum of 3 terms of an a.p. is 24 and their product is 440 the product of first and third term is

Last updated at Dec. 8, 2016 by Teachoo

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Misc 2 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers. Let the three numbers in A.P. be a d, a, a + d. Sum of the three numbers is 24 (a d) + (a) + (a + d) = 24 a + a + a + d d = 24 3a + 0 = 24 3a = 24 a = 24/3 a = 8 Also, it is given that product of three number is 440 i.e. (a d) (a) (a + d) = 440 (a d)(a + d)a = 440 (a2 d2)a = 440 Putting a = 8 (82 d2)8 = 440 82 d2 = 440/8 64 d2 = 55 64 55 = d2 9 = d2 d2 = 9 d = 9 d = 3 So, d = 3 , d = 3 When d = 3 & a = 8 The numbers are a d = 8 3 = 5 a = 8 a + d = 8 + 3 = 11 Hence the numbers are 5, 8, 11 When d = -3 & a = 8 The number are a d = 8 (-3) = 8 + 3 = 11 a = 8 a + d = 8 + (-3) = 8 3 = 5 Hence the numbers are 11, 8, & 5 Hence the numbers are 5, 8, 11 for a = 8, d = 3 and 11, 8, & 5 for a = 8, d = -3

Let the three numbers in A.P. be a – d, a, and a + d.

According to the given information,

(a – d) + (a) + (a + d) = 24 … (1)

⇒ 3a = 24

∴ a = 8

(a – d) a (a + d) = 440 … (2)

⇒ (8 – d) (8) (8 + d) = 440

⇒ (8 – d) (8 + d) = 55

⇒ 64 – d2 = 55

⇒ d2 = 64 – 55 = 9

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.

Text Solution

Solution : Let the three numbers in A.P be a-d,a and a+d. <br> According to th egiven information , we have <br>(a-d)+(a)+(a+d)=24 <br> `therefore` a= 8 <br> and (a-d)a(a+d)=440 <br> or (8-d)(8)(8+d)=440 <br> or `64-d^(2)=55` <br> or `d^(2)=9` <br> or `d=pm3` ltbr gttherefore, when d=3, the numbers are 5,8 and 11 and when d=-3, the number are 11,8, and 5. <br> Thus, the three numbers are 5,8 and 11.