Solve by Arithmetic Progression  1.Find the 11th term of 2,4,6…  2. Find the sum 1+3+5+…(2n -1)?

1.
First term, a = 2
Common difference, d = 4 - 2 = 2

11th term
= a + (n - 1) d
= 2 + (11 - 1) × 2
= 22

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2.
First term, a = 1
Common difference = 3 - 1 = 2
No. of terms = n

Sum of n terms
= n [2a + (n - 1) d] / 2
= n [2(1) + (n - 1) × 2] / 2
= n [2 + 2n - 2)] / 2
= n (2n) / 2
= n²

1) 2, 4, 6,..., 2n

so, 11th term is 2(11) = 22

2) ∑(2n - 1) => ∑2n - n

i.e. 2∑n - n

Now, ∑n => n(n + 1)/2

so, 2∑n => n(n + 1)

Hence, n(n + 1) - n => n²

i.e. the sum of the first n odd numbers is the nth square number.

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 1.
 The 11th term of 2, 4, 6, 8, … is 22.
 2. 
 a_n = 2 n - 1 
 1 + 3 + 5 +… + 39 = 400