The first three terms of an AP respectively are 3y – 1, 3y + 5 and 5y + 1. Then y equals - Mathematics

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The first three terms of an AP respectively are 3y – 1, 3y + 5 and 5y + 1. Then y equals:

(A) –3
(B) 4
(C) 5
(D) 2

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Solution

The first three terms of an AP are 3y – 1, 3y + 5 and 5y + 1 respectively

We need to find the value of y.
We know that if a, b and c are in AP, then:
b − a = c − b ⇒ 2b = a + c


∴2(3y+5) = 3y − 1 + 5y + 1

⇒6y + 10 = 8y

⇒10 = 8y −6y

⇒2y = 10

⇒y = 5

Hence, the correct option is C.

  Is there an error in this question or solution?
2013-2014 (March) Delhi Set 1

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