Question

The sum S of the arithmetic sequence a, a + d, a + 2d, . . . , a + (n – 1)d is given by \small S=\frac{n}{2}\left [2a+(n-1)d \right ] . What is the sum of the integers from 1 to 100, inclusive, with the even integers between 25 and 63 omitted?

Answer 1

Answer:

The required sum is 4214.

Step-by-step explanation:

We are given the following in the question:

An arithmetic sequence with first term a and common difference d.

The sum of n terms of A.P is given by

$S_n = \dfrac{n}{2}(2a + (n-1)d))$

We have to find:

$(1+2+3+...+98+99+100) - (26 + 28+...+60+62)$

First series:

$a = 1\\d = 1\\a_n = 100\\a + (n-1)d = 100\\\\n = 100\\\\S_{100} = \displaystyle\frac{100}{2}(2(1)+(100-1)1)\\\\S_{100} = 5050$

Second series:

$a' = 26\\d' = 2\\a'_n = 62\\a' + (n-1)d' = 62\\\\n = 19\\\\S'_{19} = \displaystyle\frac{19}{2}(2(26)+(19-1)2)\\\\S'_{19} = 836$

The require sum is:

$S_{100}-S'_{19} = 50505 - 836 = 4214$

Thus, the required sum is 4214.