If a1,a2,⋅⋅⋅,a15 are in A.P. and a1+a8+a15=15, then a2+a3+a8+a13+a14 equals
15
10
25
33
a1+a8+a15=3a1+21d=15 ⇒a1+7d=5 Now, a2+a3+a8+a13+a14=5a1+35d 5(a1+7d)=5×5=25