The sum of all two digit numbers which when divided by 4 leaves 1 as remainder is

The correct answer is: 1210

A progression of numbers known as an arithmetic sequence is one in which, for every pair of consecutive terms, the second number is derived by adding a predetermined number to the first one.
There are three types of progressions in mathematics. As follows:

• Arithmetic Progression (AP)
• Geometric Progression (GP)
• Harmonic Progression (HP)

In AP, we will come across some main terms, which are denoted as:

• First term (a)
• Common difference (d)
• nth Term (an)
• Sum of the first n terms (Sn)

13,17, 21,...., 97 are two-digit integers that, when divided by 4, leave 1 as the remainder. This creates an A.P. with a common difference of 4 and a first term of 13. Let n represent the A.P's phrase count.
We have the formula of nth term:
a_n = a + (n - 1) d
Applying this, we get:
97 = 13 + (n - 1)(4)
4(n - 1) = 97 - 13
n - 1 = 84/4
n = 22
We also have the formula of sum of n term:
S_n = n/2 [2a + (n - 1)d]
Applying this, we get:
S₂₂ = 222/2 [2(13) + (22 - 1)(4)]
S22 = 11 [26 + 84]
S22 = 11 x 110
S22 = 1210

Here we used the concept of Arithmetic progression to find the number of terms in the given series. When we study Arithmetic Progression, which is associated with: There are two key formulas we encounter, those were nth term of AP and sum of the first n terms. So therefore the sum is 1210.