Question

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

Answer 1

Answer:

X=118

Step-by-step explanation:

AP: 2,6,10....x

a=2  ,  d=4  ,  an=x  ,  Sn=1800

FOR n

we know that ,

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

putting the values...

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

1800 = n/2[4+4n-4]

1800 = n/2[4n]

1800 = 2n²

1800/2 = n²

√900 = n

30 = n

now...

an=a+(n-1)d

an=2+(30-1)4

an=2+(29)4

an=2+116

an=118

hence x is 118

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Answer 2

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