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0 votes
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in Mathematics by (71.2k points)

If rth term in expansion of (2x3 + 5/x2)10 is constant then r =______

(a) 6 

(b) 7 

(c) 4 

(d) 5

1 Answer

+1 vote
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Best answer

Correct option: (b) 7

Explanation:

Tr+1 = ncr an–r br 

Tr = ncr–1 an–r+1 br–1 

Here

Tr = 10cr–1 ∙ (2x3)10–r+1 ∙ (5 / x2)r–1 

givenTr is constant

hence

x30–3r+3 ∙ [1 / {x2r–2}] = 1

∴ x30–3r+3–2r+2 = x0 

∴ 35 = 5r

∴ r = 7

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