Question

ap is 1, 3, 5 ,7 ...find the 100th term​

Answer 1

$\large{EXPLANATION}$

Given :-

The A. P series are 1, 3, 5, 7 . . . .  

To find :-

100th term

SOLUTION :-

We have formula

an = a+(n-1)d

• d= common difference
• a = first term
• n = term

First we find the common difference (d)

Common difference will be the

7-5 = 2

5-3 = 2

3 - 1 = 2

So, the common difference is 2

d= 2

a = 1 (first term)

n = 100

We need to find an

So,

an = a+(n-1)d

an = 1+(100-1)2

an = 1+(99) 2

an = 1+198

an = 199

So, the 100th term is 199

__________________________

Know more :-

A.P means Arthemetic progression That means it follows a certain pattern that is The given sequence common difference should be same

Examples :-

2, 4 , 6 , 8 , 10. . . .

It is a Arthemetic progression beacuse its common difference is same i.e

10- 8 = 2

8 - 6 = 2

6 - 4 = 2

4 - 2 = 2

If you observe the difference that is 2 ,Since it is in A.P .

2, 4 , 6 , 8 , 10 .. . In this progression,

a = 2 [first term]

d = 2[common difference]

n = nth term

For finding nth term we have formula

an = a+(n-1) d

For finding sum of the terms we have formula that is

Sn = n/2 × 2a+(n-1)d

Answer 2

{\huge{\bold{\underline{\red{Answer}}}}}

${\huge{\bold{\underline{\green{Answer}}}}}$

$\huge\sf{Given :-}$

${ AP :- 1 , 3 , 5 , 7}$

${Here, \: first \: term \: (a) \: = 1}$

${Common \: difference \: (d) = 2}$

${To \: find \: :- \: 100th \: \: term \: }$

${i.e \: \ \: \: \: a100 \: \: = \: ?}$

${an = a + ( n-1) d}$

${a100 = 1 + ( 100 - 1)× (2)}$

${a100 = 1 + 99×2}$

${a100 = 1+198}$

${a100= 199}$

${therefore, \: 100th \: term \: is \: 199}$