Question

pls answer rrrrr helpo​

Answer 1

Given :-

First term (a)= 3.5

common difference (d)= 0

number of term (n)= 101

To find :-

The last term of Ap $\sf a_n$

◆ According to Formula

$\bigstar{\boxed{\sf {a_n=a+(n-1)d}}}$

• Where

$\longmapsto\sf a_n= last\ term\\ \\ \longmapsto\sf a= first \ term\\ \\ \longmapsto\sf d= common \ difference \\ \\\longmapsto\sf n= number \ of \ terms$

Put the values In the formula

$\longmapsto\sf a_n= a+(n-1)d\\ \\ \longmapsto\sf a_n= 3.5+(101-1)d\\ \\ \longmapsto\sf a_n= 3.5+100\times 0\\ \\ \longmapsto\sf a_n= 3.5+0\\ \\\longmapsto\sf a_n=3.5$

$\boxed{\sf{\dag{ last\ term (a_n)= 3.5}}}$

$\sf{\underline{\bigstar{Your \ answer \ is \ Option \ b}}}$

Answer 2

Arithmetic progression (AP) :-

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

___________________

Solution :-

→ First term = a = 3.5

→ common difference = d = 0

→ Number of terms = 101 .

→ Tₙ = a + (n-1)•d

Putting values we get :-

➻ Tₙ = 3.5 + (101 - 1) * 0

➻ Tₙ = 3.5 + 100 * 0

➻ Tₙ = 3.5 + 0

➻ Tₙ = 3.5 (Option B) (Ans.)