Sequence

[sidnim]

Explanations por favor, el señor!

[aashu_79]

Terms form three arithmetic progressions -

a1, a4,... nth term an = a1 + (n-1)d = 104 + (n-1)*16

a2, a5,.. nth term an = 110 + (n-1)*16

a3, a6,.. nth term an = 116+ (n-1)*16


So if (an-104) or (an-110) or (an-116) is divisible by 16, an will be included in the series.

start from 1st choice. you'll find (430 -110) is divisible by 16 so 430 is the answer.

Any short methods??

[thankont]

Nice work aashu
we can write
an = 104 +(n-1)*16 = 16*6 + (n-1)*16 + 8
an = 110 +(n-1)*16 = 16*6 +(n-1)*16 +14
an =116 +(n-1)*16 = 16*7 +(n-1)*16 + 4
so we need only to check the remainders of the division of
the numbers with 16 which can either be 4,8,14
430 leaves remainder 4

[sidnim]

simply faaaaabulous work guys...the key here is to realise that there are 3 series in one and then use the formula on those individual series...i stumbled on the first hurdle & didnt realise this fact

[aashu_79]

Nice work aashu
we can write
an = 104 +(n-1)*16 = 16*6 + (n-1)*16 + 8
an = 110 +(n-1)*16 = 16*6 +(n-1)*16 +14
an =116 +(n-1)*16 = 16*7 +(n-1)*16 + 4
so we need only to check the remainders of the division of
the numbers with 16 which can either be 4,8,14
430 leaves remainder 4

Thats better. Thanks buddy

[MikeJung]

an = 104 +(n-1)*16 = 16*6 + (n-1)*16 + 8
an = 110 +(n-1)*16 = 16*6 +(n-1)*16 +14
an =116 +(n-1)*16 = 16*7 +(n-1)*16 + 4

sorry but how did u come up with those equation?
thanks in advance.