Subjective Type

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

Solution

Substituting $$n= 1, 2, 3, 4, 5$$ we obtain
$$\displaystyle { a }_{ 1

}=1.\frac { { 1 }^{ 2 }+5 }{ 4 } =\frac { 6 }{ 4 } =\frac { 3 }{ 2 } \\ {

a }_{ 2 }=2.\frac { { 2 }^{ 2 }+5 }{ 4 } =2.\displaystyle \frac { 9 }{ 4 } =\frac { 9

}{ 2 } \\ { a }_{ 3 }=3.\displaystyle \frac { { 3 }^{ 2 }+5 }{ 4 } =3.\frac { 14 }{ 4

} =\frac { 21 }{ 2 } \\ { a }_{ 4 }=4.\displaystyle \frac { 4^{ 2 }+5 }{ 4 } =21\\ { a

}_{ 5 }=5.\displaystyle \frac { { 5 }^{ 2 }+5 }{ 4 } =5.\frac { 30 }{ 4 } =\frac { 75

}{ 2 } $$

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 _______?

Sequences and Series

$$\displaystyle { a }_{ n }=n(n+2)$$

Sequences and Series

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

Sequences and Series

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

Sequences and Series

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

Sequences and Series

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

Sequences and Series

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

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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