Question

Find the value of x when in the A.P given below 2+6+10+....x=1800

Answer 1

answer : x = 118

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

here ; 2, 6, 10 , .....x are in arithematic progression where 2 is first term and 4 is common difference of ap.

use formula, $T_n=a+(n-1)d$ to find number of terms in ap.

x = 2 + (n - 1) × 4

x = 2 + 4n - 4

x = 4n - 2

x + 2 = 4n

n = (x + 2)/4

now, use formula, $S_n=\frac{n}{2}(a+T_n)$

here, $S_n=1800,n=\frac{x+2}{4},a=2,T_n=x$

1800 = {(x + 2)/4}/2 [2 + x ]

1800 = (x + 2)/8 × (x + 2)

1800 × 8 = (x + 2)²

14400 = (x + 2)²

(120)² = (x + 2)²

x + 2 = 120 => x = 118

hence, value of x = 118

Answer 2

Answer:

118 = X

Step-by-step explanation: Hey mate,here you go....

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..

Hope this helps you and plz mark me as brainliest.

▪︎■♡■▪︎♡■▪︎♡■▪︎