Question

is 100 a term of -1,3,7,11...?
pls give .e answer as fast as possible​

Answer 1

Answer:

an=a+(n-1)d

-1+(100-1)4

-1+99×4

-1+396

100term =395

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Answer 2

Here,

AP : { -1, 3, 7, 11 ......}

$\rm{}a_{ \tiny{1}} = - 1 \\ \rm{}d = a_{ \tiny{n}} - a_{ \tiny{(n - 1)}} = 4$

Now,

$\text{We'll find whether 100 is} \: \: \rm{n}^{th} \: { term} \\ \text{of the AP. So, If n is positive integer, then,} \\ \text{ 100 can be one \: of \: the terms of AP.}$

Let's

$\text{Assume 100 as} \: \: {\rm{ a_{n}}} \: \:$

Then, By Formula of finding nth terms,

$\rm{}a_n= a + (n - 1)d \\ \bf = > 100 = (- 1) + (n - 1)4 \\ \bf = > 100 = - 1 + 4n - 4 \\ \bf = > 100 = 4n - 5 \\ \bf = > 100 + 5 = 4n \\ \bf = > n = \frac{105}{4} = 26.25$

Since,

$\rm{}n = 26.25 \: \: which \: is \: not \: an \: integer.$

Thus,

• 100 is not the term of AP.