the value of 4C3+5C3+6C3...12C3 is equal to??​

  • matematika

    Subject:

    Math
  • Author:

    penny
  • Created:

    1 year ago

Answers 1

Step-by-step explanation:

Given→ AP→4C3+5C3+6C3...12C3

Solution→ a=4C3 d=a2 - a1 n=?

=5C3 - 4C3

=1C3

12C3=a+(n-1)d

12C3=4C3+(n-1)1C3

12C3 - 4C3=(n-1)1C3

8C3=(n-1)1C3

8C3/1C3=n-1

8=n-1

8+1=n

9=n

So, number of terms are 9

S=n/2[2a+(n-1)d]

S=9/2[2*4C3+(9-1)1C3]

S=9/2[8C3+(8)1C3]

S=9/2[8C3+8C3]

S=9/2[16C3]

S=9[8C3]

S=72C3

So,the answer is 72C3.

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