If and are in AP.Then, find the value of 
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ⁿC₁ , ⁿC₂ & ⁿC₃ are in AP
=> 2 * ⁿC₂ = ⁿC₁ + ⁿC₃
=> 2 * n!/((n-2)!2!) = n!/(n-1)!1! + n!/(n-3)!3!
=> 2 * n(n-1) / 2 = n + n(n-1)(n-2)/6
=> 6n(n-1) = 6n + n(n-1)(n-2)
=> 6(n-1) = 6 + (n-1)(n-2)
=> 6n - 6 = 6 + n² - 3n + 2
=> n² - 9n + 14 = 0
=> n² - 2n - 7n + 14 = 0
=> n(n-2) - 7(n-2) = 0
=> (n-7)(n-2) = 0
=> n = 7 n can not be less than 3 so n = 2 is not possible
⁷C₁ + ⁷C₂ + ⁷C₃
= 7 + 21 + 35
= 63
hope this helps you.
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Answer:
63 is answer dear friend
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