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There are two Arithmetic Progressions A and B such that their nth terms are given by An = 101 + 3(n 1) and Bn = 150 + (n 1), where n is the set of natural numbers. The first 50 terms of A and B are written alternately i.e. A1B1A2B2..A50B50. What is the remainder when the number so formed is divided by 11?...

what can be the coolest approach to solve such problems...:-P

In such problems drz alwys sm kinda pattern.

here all d terms are of 3 digits and acc to div rule of 11 jst subtract the triplets adjacent to each odr and d diff will gv d remainder.

In dis q a1=101, b1=150, b1-a1=49, rem=5

a2=104, b2=151, b2-a2=47, rem=3

a3=107, b3=152, b3-a3=45, rem=1

a4=110, b4=153, b4-a4=43, rem= -1

continue in dis manr nd u will find d 1st 11 rem will sum out to b 0.

so for num till a44b44, rem wil b 0

nw we need to find the rem for rest of terms from a45,

a45=233, b45=194, diff=39, rem= -5

Here no need to solve further as u the pattern will b familier, the pattern of rem wil b lyk dis

-5, -3, -1, 1, 3 ,5

only 6 terms were left frm a45 to a50, so the will give the above rem.. and hence the total rem wil b zero.

dis approach is more of reasoning, so u need to give it a try by urself 4 its cmplt understanding...

Mine is coming 43.

Have u taken care that the pairs give birth in the 2nd month.

Have u taken care that the pairs give birth in the 2nd month.

Mine is coming 43.

Have u taken care that the pairs give birth in the 2nd month.

Have u taken care that the pairs give birth in the 2nd month.

66 pairs ? is it ? correct me pls

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