Assertion :If ${^{100} C}_{r}, {^{100} C}_{r + 1}, {^{100} C}_{r + 2}, {^{100} C}_{r + 3}$ are in A.P, then $r = 49$ Reason: Four consecutive coefficients of a binomial can never be in A.P.

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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Assertion is correct but Reason is incorrect

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Assertion is incorrect but Reason is correct

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Solution

The correct option is D Assertion is incorrect but Reason is correct
$^{100} C_{r},^{100} C_{r + 1},^{100} C_{r + 2}$ are in A.P.
Replacing $r = a - 1$ we get
$^{100} C_{a - 1},^{100} C_{a},^{100} C_{a + 1}$ are in A.P.
Then it must follow
$(n - 2 a)^{2} = n + 2$
$(100 - 2 a)^{2} = 102$
102 is not a perfect square.
Hence a will be irrational real number.
However a is needed to be a whole number.
This is contradictory.
Hence the above terms are not in A.P.

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Assertion :If 100Cr,100Cr+1,100Cr+2,100Cr+3 are in A.P, then r=49 Reason: Four consecutive coefficients of a binomial can never be in A.P.

Assertion :If ${^{100} C}_{r}, {^{100} C}_{r + 1}, {^{100} C}_{r + 2}, {^{100} C}_{r + 3}$ are in A.P, then $r = 49$ Reason: Four consecutive coefficients of a binomial can never be in A.P.