Question

(ii) Find the sum of all numbers of two digits which leaves the remainder 1
when divided by 4.​

Answer 1

Answer:

Step-by-step explanation:

The first two digit number which when divided by 44 leaves remainder 11 is 43+1=1343+1=13 and last is 424+1=97424+1=97; form (4k+1)(4k+1).

Thus we have to find the sum of 13+17+21+13+17+21+______+97+97,

which is an A.P. \therefore 97=13+(n-1)\cdot 4∴97=13+(n−1)⋅4

\therefore n=22∴n=22

\therefore S=(n/2)[a+l]=11(13+97)∴S=(n/2)[a+l]=11(13+97)

=11\times 110=1210=11×110=1210.

HOPE it will help you and please mark it as brainliest

Answer 2

Answer:

1210

Step-by-step explanation:

let n be the number of 2 digit number when divided by 4 leaves remainder 1

that means n-1 is divisible by 4

that is ,n-1=4k,k is a integer

n=4k+1

according to condition ,

9<n<100

9<4k+1<100

8<4k<99

therefore ,k=3,4,5,6,7......24

n=13,17,21,25,29,33,37,.........,97

Sn=22/2[13+97]

=11(110)

= 1210