Question

Correctly explained answer to be BRAINLIEST. ​

Answer 1

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 2

Answer:

$\huge\underline\bold\red{AnsWeR}$

$= 7 + 27 + 35 \\ \\ = 65 \\ \\ \\ is \: your \: answer$