Q). Find the value of X, when in the A.P. given below 2 + 6 + 10 + …… + X = 1800
Ans:
Step 1: In the given AP: a = 2, d = 4,
Let’s consider there are n terms in this AP,
Given that the sum of AP is 1800 ∴ Sum of first n terms of AP is 1800.
Since X is the last term of AP ∴ X is the nth term and we need to find value of nth term.
Step 2: Since Sn = 
[2 a + (n – 1)d]
[2 a + (n – 1)d]∴ 1800 = [2 (2) + (n – 1)(4)]
[2 (2) + (n – 1)(4)]∴ 1800 = (n)[2 + (n – 1)(2)]
∴ 1800 = (n)(2 + 2 n – 2)
∴ 1800 = (n)(2 n)
∴ 900 = n2
∴ n2 = 302
∴ n = 30
It means there are 30 terms in the given AP.
Step 3: Now value of nth term, Tn = a + (n – 1) d
∴ T30 = (2) + (30 – 1)(4)
∴ T30 = 2 + 29 x 4
∴ T30 =118
Therefore, the value of X is 118.
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