Hint: We need to find the sum of two digit numbers of the form 4k +1. We can see that these numbers form an arithmetic sequence with a common difference d. So to find the sum we need to find the number of terms using the formula an=a+(n−1)d
an=a+(n−1)d and then find the sum using the formula Sn=n/2(a+an)Sn=n/2(a+an)
Complete step-by-step answer:
We are asked for sum of two digit numbers which leaves a remainder 1 when divided by 4
Hence the numbers must be a successor of numbers which are multiple of four
We need to find the sum of two digit numbers of the form 4k + 1 , where
Step 2 :
So from this we get the numbers to be 13 , 17 , 21 , …… , 97
We can see that there exist a common between the numbers
(i.e.) 17 – 13 =4
21 – 17 =4
Therefore the numbers form an A.P with a common difference , d = 4
Step 3 :
In order to find the sum of these numbers we need to find the number of terms in this sequence.
Here , we have the first term to be 13 and the last term to be 97
(i . e.)a=13 and an=97a=13 and an=97
Since an=97 an=97, we know that the formula to find the nth term of the sequence is
⇒an=a+(n−1)d⇒an=a+(n−1)d
Lets substitute the known values in the above formula
⇒97=13+(n−1)4
⇒97−13=4n−4
⇒84=4n−4
⇒84+4=4n
⇒88=4n
⇒88/4=n
⇒n=22
Hence there are 22 terms in the sequence.
Step 4 :
Now let's find the sum of the terms of the sequence.
Sum of the terms is given by the formula
⇒Sn=n/2(a+an)⇒Sn=n/2(a+an)
We have n = 22 and a = 13 and an=97an=97
Lets substitute it in the above formula
⇒S22=22/2(13+97)
⇒S22=11(110)
⇒S22=1210
Therefore the sum of two digit numbers which when divided by 4 leaves a remainder 1 is 1210.
Note: The above formula for the sum of n terms can be used only if we know the last term of the sequence or series , when we do not know the last term we can use this alternate formula
⇒Sn=n/2(2a+(n−1)d)
