Question

find the value of x when in the ap given below . 2+6+10+........+x=1800​

Answer 1

a=2,d=4

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

Sn=1800

therefore ,

1800=n/2(2*2+(n-1)4)

1800=n/2(4+4n-4)

3600=n(4n)

3600=4n^2

3600/4=n^2

900=n^2

√900=n

n=30

there a30=x

a30=2+(30-1)4

a30=2+29*4

a30=118=X..

Answer 2

Answer:

The value of x is 118.

Step-by-step explanation:

we have to find the value of x when in AP

2+6+10+........+x=1800​

First term=a=2

Common difference=d=4

As the sum formula for A.P is

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_n=1800$

∴ $1800=\frac{n}{2}(2(2)+(n-1)4)$

$1800=\frac{n}{2}(4+4n-4)$

$3600=n(4n)$

$3600=4n^2$

$\frac{3600}{4}=n^2$

$900=n^2$  

$n=\sqrt{900}=30$

Hence, $a_{30}=x$

$x=2+(30-1)4$

$x=118$

The value of x is 118.