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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Binomial Theorem

BITSAT 2020 · Mathematics · Binomial Theorem

If a, b, c are three natural numbers in AP and a + b+c=21 then the...

If a, b, c are three natural numbers in AP and a + b+c=21\mathrm{b}+\mathrm{c}=21 then the possible number of values of the ordered triplet (a,b,c)(\mathrm{a}, \mathrm{b}, \mathrm{c}) is

Options

  1. A15
  2. B14
  3. C13
  4. DNone of these

Correct answer

C. 13

Step-by-step solution

\quad Let a=b−dandc=b+da=b-d and c=b+d, then a+b+c=21⇒b=7a+b+c=21\Rightarrow \mathrm{b}=7 So, the equation is a+c=14∴\mathrm{a}+\mathrm{c}=14\therefore No. of solution = coeff. of x14in(x+x2+….)=13C12=13x^{14} in \left(x+x^{2}+\ldots .\right)={ }^{13} \mathrm{C}_{12}=13
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