Subjective Type

$$\displaystyle{ a }_{ n }={ (-1) }^{ n-1 }\quad { 5 }^{ n+1 }$$

Solution

Given, $$a_n=(-1)^{n-1}. 5^{n+1}$$
Substituting $$n= 1, 2, 3, 4, 5$$, we obtain
$$\displaystyle { a }_{ 1

}={ (-1) }^{ 1-1 }{ 5 }^{ 1+1 }={ 5 }^{ 2 }=25\\ { a }_{ 2 }={ (-1) }^{

2-1 }{ 5 }^{ 2+1 }={ -5 }^{ 3 }=-125\\ { a }_{ 3 }={ (-1) }^{ 3-1 }{ 5

}^{ 3+1 }={ 5 }^{ 4 }=625\\ { a }_{ 4 }={ (-1) }^{ 4-1 }{ 5 }^{ 4+1 }={ 5

}^{ 5 }=-3125\\ { a }_{ 5 }={ (-1) }^{ 5-1 }{ 5 }^{ 5+1 }={ 5 }^{ 6

}=15625$$

SIMILAR QUESTIONS

Sequences and Series

Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than $$200$$ and less than $$220$$. If the second term in it is $$12$$, then $$4^{th}$$ term is :

Sequences and Series

Let X be the set consisting of the first $$2018$$ terms of the arithmetic progression $$1, 6, 11,$$______, and Y be the set consisting of the first $$2018$$ terms of the arithmetic progression $$9, 16, 23$$, ________. Then, the number of elements in the set X$$\cup$$Y is _______?

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$$\displaystyle { a }_{ n }=\frac { n }{ n+1 } $$

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$$\displaystyle{ a }_{ n }={ 2 }^{ n }$$

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$$\displaystyle{ a }_{ n }=\frac { 2n-3 }{ 6 } $$

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$$\displaystyle{ a }_{ n }=n\frac { { n }^{ 2 }+5 }{ 4 } $$

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$$\displaystyle{ a }_{ n }=4n-3;{ a }_{ 17 },{ a }_{ 24 }$$

Sequences and Series

$$\displaystyle { a }_{ n }=\frac { { n }^{ 2 } }{ { 2 }^{ n } } ;{ a }_{ 7 }$$

Sequences and Series

$$\displaystyle{ a }_{ n }=({ -1 })^{ n-1 }{ n }^{ 3 };{ a }_{ 9 }$$

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