Q38 of 40 Page 1

The sum of four consecutive numbers in AP is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7: 15 . Find the numbers.

OR


Solve: 1+4+7+10+……+x =287


Let the four conductive terms of AP be

(a - 3d), (a - d) , (a + d) , (a + 3d)


Case I : a – 3d + a – d + a + d + a + 3d = 32


4a = 32


a = 8


CASE II :



When d = 2 & a = 8


a – 3d = 8 – 3(2) = 2


a – d = 8 – 2 = 6


a + d = 8 + 2 =10


a + 3d = 8 + 3(2) = 14


when d = -2 & a = 8


a - 3d = 8 – 3(-2) = 14


a - d = 8 – (-2) = 10


a + d = 8 + (-2) = 6


a + 3d = 8 + 3(-2) = 2


OR


Let the number of terms be ‘n’


Given, sum = 287




⇒ n(2 + 3n – 3) = 574


⇒ n(3n – 1) = 574


⇒ 3n2 – n – 574 = 0


⇒ 3n2 – 42n + 41n – 574 = 0


⇒ 3n(n – 14) + 41(n – 14) = 0


⇒ (3n + 41)(n – 14) = 0


⇒ n = 14 or -41/3 [Not possible]


Now, 14th term = a + 13d = 1 + 13(3) = 1 + 39 = 40


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