Question

If A1,A2,A3,...... is an Arithmetic progression with common difference 1 and A1+A2+A3+A4+.......+A98=137 then A2+A4+A6+.....+A98=

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

A₂ +A₄ +A₆  ... + A₉₈  = 93    

Step-by-step explanation:

Given data

A₁ A₂ A₃ ... A₉₈ is an Arithmetic progression with common difference 1 `

⇒  A₁+A₂+A₃+A₄+ ...+ A₉₈ = 137 _(1)  

here we need to find A₂+A₄+A₆ + ...  + A₉₈  

the common difference is 1 then

⇒  A₂-1 = A₁   A₄-1 = A₃ ...  

(1) ⇒ (A₂-1) + A₂ +(A₄-1) + A₄+  ...+ A₉₈ = 137  

        A₂ - 1 + A₂ + A₄ - 1 + A₄ +  ...+ A₉₈ = 137  

        (A₂ +A₂ +A₄ +A₄ ..) - 49  = 137

        2A₂ + 2A₄ + 2A₆ ..+2A₉₈  = 137 + 49

        2( A₂ +A₄ +A₆  ... + A₉₈) = 186

        A₂ +A₄ +A₆  ... + A₉₈   = 186 / 2 =93

       A₂ +A₄ +A₆  ... + A₉₈  = 93