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Question

If the 6th, 7th and 8th terms in the expansion (x+a)n are respectivley 112, 7 and 14, find x,a,n.


Solution

It is given that,

T6=112,T7=7,T8=14

T6=nCn5xn5×a5=112

T7=nCn6xn6×a6=7

and,T8=nCn7xn7×a7=14

Now,

T7T6=nCn6xn6×a6nCn5xn5×a5=7112

nCn6nCn5×ax=116n6+1n(n5)+1×ax=116

[nCrnCr1=nr+1r] n56×ax=116

ax=616×1n5

ax=38×1(n5)....(i)

and,

T8T7=nCn7xn7×a7nCn6xn6×a6=147

T8T7=nCn7nCn6=ax=128

nCn7nCn6×ax=128

n7+1n(n6)+1×ax=128

n67×ax=128

ax=14(n6)....(ii)

Comparing equation (i) and (ii), we get 

38×1(n5)=14(n6)

32×1(n5)=1(n6)

3(n6)=2(n5)

3n18=2n10

3n2n=1810

n=8

Putting in equation (i) we get

a=x

Now, 8C5x85(x8)5=112

=56x885=112

x8=48

x=4

By putting the value of x and n in (i) we get

a=12

a=12 and x=4

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