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Question

# If the 2nd,3rd and 4th terms in the expansion of (x+a)n are 240,720 and 1080 respectively, then the value of least term in the expansion is

A
16
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B
32
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C
64
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D
81
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Solution

## The correct option is B 322nd term is nC1xn−1a=240⋯(1) 3rd term is nC2xn−2a2=720⋯(2) 4th term is nC3xn−3a3=1080⋯(3) Now, multiplying equations (1) and (3) and dividing by the square of equation (2), we get nC1×nC3(nC2)2=240×1080(720)2⇒n×n(n−1)(n−2)(2!)2n2(n−1)2×3!=12⇒(n−2)×4=3×(n−1) (∵n≠1) ⇒n=5 Putting n=5 in (1) and (2), we get 5x4a=240 and 10x3a2=720 ⇒(5x4a)210x3a2=(240)2720⇒x5=32∴x=2∴a=2405x4=4824=3 Now, (2+3)5=25+ 5C124×3+ 5C223×32+ 5C322×33+ 5C42×34+ 5C535=32+240+720+1080+810+243 Hence, least value term in the expansion is 32.

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