Find the sum of all 11 term of an AP whose middle most term in 30.

1.	Find the sum of all 11 term of an AP whose middle most term in 30.
Answer» Let ‘a’ be the first term and ‘d’ be the common difference of the give A.P. Middle most term = (11 + 1/2)^th = 6^th term t_n = a + (n – 1)d t₆ = a + 5d a + 5d = 30 S_n = (n/2) [2a + (n – 1)d] S₁₁ = (11/2) [2a + 10d] = (11/2) x 2 [a + 5d] = 11 x 30 = 330 Sum of 11 terms = 330

Answer»

Let ‘a’ be the first term and ‘d’ be the common difference of the give A.P.

Middle most term = (11 + 1/2)^th = 6^th term

t_n = a + (n – 1)d

t₆ = a + 5d

a + 5d = 30

S_n = (n/2) [2a + (n – 1)d]

S₁₁ = (11/2) [2a + 10d]

= (11/2) x 2 [a + 5d]

= 11 x 30

= 330

Sum of 11 terms = 330

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