Question

Find the value of X when A.P is
2+6+10.......+X =1800.
Please tell me fast

Answer 1

Answer:

118

Step-by-step explanation:

Given 2 + 6 + 10 ..... x = 1800 .

Upon observing the above sequence we note these things :-

→ The above sequence has a common difference in them .

→ The terms are increasing by 4 .

→ The common difference is hence 4 .

→ The first term is a .

→ The above series is an arithmetic progression .

n th term of an A.P is given by the formula :-

a + ( n - 1 ) d .

Sum of n terms is given by the formula :-

Sn = n/2 [ 2 a + ( n - 1 ) d ]

Now putting the above values :-

Let there be n terms .

a = 2

d = 4

Sn = 1800 .

1800 = n/2 [ 2(2) + ( n - 1 )( 4 ) ]

⇒ 3600 = n [ 4 + 4 n - 4 ]

⇒ 3600 = n × 4 n

⇒ 4 n² = 3600

⇒ n² = 3600/4

⇒ n² = 900

⇒ n = 30

Hence n = 30 .

Now n th term = x .

a + ( 30 - 1 ) d = x

⇒ a + 29 d = x

⇒ x = 2 + 29 × 4

⇒ x = 2 + 116

⇒ x = 118

∴ The value of x is 18 .

Answer 2

$\mathfrak{Answer:}$

118.

$\mathfrak{Step-by-step\:\:explanation:}$

$\underline{\bold{Important\:Formula:}}\\\\\\\bold{1.\quad T_n=a+(n-1)d}\\\\\\\bold{2.\quad S_n=\dfrac{n}{2}[2a+(n-1)d]}\\\\\\\bold{3.\quad S_n=\dfrac{n}{2}[a+l]}$

$\underline{\bold{Where,}}\\\\\\\tt{a=First\:term.}\\\\\tt{T_n=nth\:term.}\\\\\tt{l=last\:term.}\\\\\tt{S_n=sum\:of\:n\;terms.}\\\\\tt{d=common\:difference.}$

Given :

• AP = 2 , 6 , 10 , ....... , x.
• Sum of n terms = 1800.
• a = 2.
• d = 4.

$\implies\bold{S_n=\dfrac{n}{2}[2a+(n-1)d]}\\\\\\\implies\bold{1800=\dfrac{n}{2}[2 \times 2 +(n-1)4]}\\\\\\\implies\bold{1800=\dfrac{n}{2}[4+4n-4]}\\\\\\\implies\bold{1800=\dfrac{4n^2}{2}}\\\\\\\implies\bold{n^2=\dfrac{1800\times 2}{4}}\\\\\\\implies\bold{n^2=900}\\\\\\\implies\bold{n=\pm 30.}\\\\\\\underline{\bold{Possible\:value\:of\:n=+30.}}\\\\\\\tt{So}\\\\\\\bold{T_{30}=x=a+(n-1)d}\\\\\\\tt{=2+(30-1)4}\\\\\\\tt{=2+29\times 4=2+116=118.}$

$\boxed{\boxed{\bold{T_{30}=x=118.}}}$