If A = (x–a) (x–b) (x–c) .............(x–z). Then number of terms in the expansion of (a+A) (b+A) (c+A) ..........(z+A) is

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Solution

With a little bit zooming we see prior to (x-z) there is (x-y ) and just before (x-y) there is (x -x) which is equals to 0 and hence product of entire sequence will be zero (0). So the value of A = (x-a)(x-b)(x-c)…..(x-z) is 0.
Thus, the expansion of the term (a+A)(b+A)........(z+A) will be abc............z as A = 0.
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