If the middle term in the expansion of $$x2+28is$$ $$1120$$, then the product of possible real values of x is $$____$$.

Middle term is $$n2+1th$$ , i.e., $$(4+1)th$$ , i.e., $$T5$$∴ $$T5=8C4x24$$×$$24=1120or$$ $$x4=8$$×$$7$$×$$6$$×$$51$$×$$2$$×$$3$$×$$4x4=1120$$⇒$$x4=112070=16$$⇒$$x2+4x2$$−$$4=0Therefore$$, $$x=$$±$$2$$ only as x∈R

Binomial theorem for positive integral Index

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If the middle term in the expansion of ${(\frac{x}{2} + 2)}^{8}$ is 1120, then the product of possible real values of x is ____.

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Solution

Middle term is ${(\frac{n}{2} + 1)}^{th}, i.e., (4 + 1)^{th}, i.e., T_{5}$
$∴ T_{5} =^{8} C_{4} {(\frac{x}{2})}^{4} \times 2^{4} = 1120$
or $x^{4} = \frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4} x^{4} = 1120$
$\begin{array}{l} \Rightarrow x^{4} = \frac{1120}{70} = 16 \\ \Rightarrow (x^{2} + 4) (x^{2} - 4) = 0 \end{array}$
Therefore, $x = \pm 2 only as x \in R$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If the middle term in the expansion of x2+28is 1120, then the product of possible real values of x is ____.

Middle term is n2+1th , i.e., (4+1)th , i.e., T5∴ T5=8C4x24×24=1120or x4=8×7×6×51×2×3×4x4=1120⇒x4=112070=16⇒x2+4x2−4=0Therefore, x=±2 only as x∈R

Ready to Test Your Skills?

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If the middle term in the expansion of ${(\frac{x}{2} + 2)}^{8}$ is 1120, then the product of possible real values of x is ____.