In an AP the sum of
second & third term is 22 & the
Product of first & fourth term is
85. Then find th

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In an AP the sum of
second & third term is 22 & the
Product of first & fourth term is
85. Then find the sum of first
10 terms by considering the
positive value of common
difference.​

1 thought on “In an AP the sum of<br />second & third term is 22 & the<br />Product of first & fourth term is<br />85. Then find th”

  1. Answer:

    When the consecutive terms of series differ by a common number, then the series is said to be Arithmetic Progression


    Let a be the first term of the AP


    d be the common difference of the AP


    nth term of AP ⇒ a + ( n – 1) d


    Given,


    The sum of second and third term is 22


    ⇒ (a + d) + (a + 2d) =22


    ⇒ 2a + 3d = 22


    ⇒ d = 1/3 ( 22 – 2a)


    The product of first and fourth term is 85


    ⇒ a ( a + 3d) = 85


    ⇒ a² + 3ad = 85


    Substituting the value of d gives,


    ⇒ a ( a + 3 ( 1/3 * (22 – 2a) )) = 85


    ⇒ a ( a + 22 – 2a) = 85


    ⇒ a ( – a + 22) = 85


    ⇒ – a² + 22a = 85


    ⇒ a² – 22a + 85 = 0


    ⇒ a² – 17a – 5a + 85 = 0


    ⇒ a ( a – 17) – 5 ( a – 17)= 0


    ⇒ (a-5)(a-17)= 0


    ⇒ a = 5 or a = 17


    If a = 5,


    d = 1/3 ( 22 – 10) = 1/3 ( 12) = 4


    If a = 17,


    d = 1/3 ( 22 – 34) = 1/3 ( – 12) = – 4


    a = 5, d = 4


    Then Arithmetic Progression is


    5, 9, 13, 17


    a = 17, d = – 4


    Then Arithmetic Progression is,


    17, 13, 9, 5


    Therefore, The required terms in the AP are 5, 9, 13, 17.

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