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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 32
2nd term is nC1xn1a=240(1)
3rd term is nC2xn2a2=720(2)
4th term is nC3xn3a3=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)2n×n(n1)(n2)(2!)2n2(n1)2×3!=12(n2)×4=3×(n1) (n1)
n=5

Putting n=5 in (1) and (2), we get
5x4a=240 and 10x3a2=720
(5x4a)210x3a2=(240)2720x5=32x=2a=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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