Question

find the value of X. for the given A.P

$2 + 6 + 10 + ....... + x = 1800$
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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, $S_n=1800,n=\frac{x+2}{4},a=2,T_n$

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