Question

Find the sum of all the 11 terms of an A.P. whose middle most term is 30.

Answer 1

Since, the total number of terms (n) = 11  ...[odd]

∴ Middle most term = `(n + 1)^("th")/2` term

= `((11 + 1)/2)^("th")` term

= 6^th term

Given that,

a₆ = 30  ...[∵ a_n = a + (n − 1)d]

⇒ a + (6 − 1)d = 30    

⇒ a + 5d = 30    ...(i)

∵ Sum of n terms of an AP,

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

∴ S₁₁ = `11/2[2a + (11 - 1)d]`

= `11/2(2a + 10d)`

= 11(a + 5d)

= 11 × 30   ...[From equation (i)]

= 330