Question

The sum of 1st 11terms of an AP whose middle term is 30?

Answer 1

Tn = a+(n-1)d

T6 = a+5d = 30 ---------------------- (1)

(1) *2 => 2a+10d = 60

Sn = $\frac{n}{2}$[2a+(n-1)d]

S11 = $\frac{11}{2}$[2a+10d]

     = $\frac{11}{2} [60]$

     = 11*30 = 330

Answer 2

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