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Deepak Bora
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If the sum of the First n term of an AP is given by S_n=(3n^2-n), Find its (i)nth term (ii) first term (iii) common difference.

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ICSE & CBSE Board Question Based on Arithmetic Progression of RS Aggarwal
Here the sum of the First n terms of AP is given
You have to find
(i)nth term
(ii)first term
(iii)common difference
This is the Question Number 25 Of Exercise 11 C of RS Aggarwal Solution.

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  1. Let to be first term ‘a’ and it’s CommOn difference ‘d’ and it’s nth terms ‘nth’.
    Given;- S_n = ( 3n^2 – n ).
    Let to be n replace (n-1)
    °•° S_n=3n²-n
    S_n-1= 3(n-1)²-(n-1)
    = 3(n²-2n+1-(n-1)
    = 3n²-6n+3-n+1
    = 3n²-7n+4
    Now, According to the Question statement;-
    T_n=S_n-S_n-1
    = 3n²-n-(3n²-7n+4)
    = 3n²-n-3n²+7n-4
    =6n-4
    Hence, It’s Nth term= 6n-4
    2) Substitute 1 in Equation as replace n
    => 6n-4
    => 6(1)-4
    = 6-4
    => 2
    Hence, 2 is the first term of this Arithmetic Sequence or Progression Equation!-
    3) Substitute 2 in Equation as replace n
    => 6(2)-4
    => 12-4
    => 8
    Hence, Arithmetic Sequence:- 2, 8
    Now, Considering on Question as Third follow;-
    CommOn Difference = T_2-T_1
    = 8-2
    = 6
    Conclusion;-
    i ) nth term= 6n-4
    ii ) first term = 2
    iii) Common difference. = 6

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