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If p is a real number and if the middle term in the expansion of (p2+2)8 is 1120, find p.

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Question

If p is a real number and if the middle term in the expansion of `(p/2 + 2)^8` is 1120, find p.

Sum
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Solution

Given expression is `(p/2 + 2)^8` 

Number of terms = 8 + 1 = 9 (odd)

∴ Middle term = `(9 + 1)/2` th term = 5th term

∴ T5 = T4+1

= `""^8"C"_4 (p/2)^(8 - 4)  (2)^4`

= `""^8"C"_4  p^4/2^4 xx 2^4`

= `""^8"C"_4 p^4`

Now 8C4P4 = 1120

⇒ `(8 xx 7 xx 6 xx 5)/(4 xx 3 xx 2 xx 1) "P"^4` = 1120

⇒ 70P4 = 1120

⇒ P4 = `1120/70` = 16

⇒ P4 = 24

⇒ P = ±2

Hence, the required value of P = ±2

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Chapter 8: Binomial Theorem - Exercise [Page 143]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 8 Binomial Theorem
Exercise | Q 12 | Page 143

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