Question

Find the sum of all the 11 terms of an AP 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

∵ Sum of n terms of an AP

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

∴ `S_11 = 11/2[2a + (11 - 1)d]`

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

= `11(a + 5d)`

= `11 xx 30`

= 330  .....[From equation (i)]