Question

Q:1-The sum of all integers
between 2 and 100 which are
divisible by 3 is...​

Answer 1

The following Question is a part of Arithmetic Progression.

Answer is 1683.

To find:-

• The sum of all integers  between 2 and 100 which are  divisible by 3.  

Solution:-

If the number must be divisible by 3, so first number between 2 and 100 is 3, and the last number is 99.

How the last number is 99?

$\tt 100 \div 3 = 33 \times 3 + \hl{1} \\\\ \tt Here\:\:1 \:\:is \:\:left\:\:as \:\:remainder \\\\ \tt So\:\:subtract\:\:1\:\:from \:\:100$

We get answer as 99. This is how we choose the last term as 99

Now we have

$\tt \star a=3 \\\\ \tt \star a_n=99 \\\\ \tt \star d = 6-3 = 3$

So, we have to find number of terms first

$\tt a_n = a+(n-1)d \\\\ \tt \Rightarrow 99 = 3 +(n-1)3 \\\\ \tt \Rightarrow 96 = (n-1)3 \\\\ \tt \Rightarrow \dfrac{96}{3}=n-1 \\\\ \tt \Rightarrow 32 = n-1 \\\\ \tt \Rightarrow 33 = n$

So the number of terms are 33

Now finding $\tt S_n$

$\tt S_n = \dfrac{n}{2}[a + a_n] \\\\ \tt = \dfrac{33}{2}[3+99] \\\\ \tt =\dfrac{33}{2}[102] \\\\ \tt = 33[51] \\\\ \tt = 1683$

So, the sum of the numbers are 1683

Answer 2

Answer:

The series of the positive odd integers  between 2 and 100 that are divisible by e are:

3, 9, 15  21, 27 ..................99

a1 = 3, a2 = 9, a3 = 21

common difference (d) = a2-a1  

                                      = 9-3

                                      = 6

again d= a3-a2

           =21 -15

           =6

3, 9, 15  21, 27 ..................99 is a series of Arithmetic Progression

Now,

The last term = 99

Applying the formula to find the nth term,

$a_{n}$ = a1+(n-1)

99 =3+(n-1) x 6

99-3 = (n-1) x 6

96 = 6n-6

96+6 = 6n

102 = 6n

102/6 = n

17 = n

n=17

There are 17 odd numbers divisible by 3 between 2 and 100

Now, we need to find the sum of all those 17 numbers,

so, n=17

applying the formula to find the sum of nth terms

$S_{n}$ = $\frac{n}{2}$ [ 2a + (n-1)d]

$S_{17}$ = $\frac{17}{2}$ [ 2 x 3 + ( 17-1)6]

$S_{17}$ = $\frac{17}{2}$ [6+16 x 6]

$S_{17}$ =  $\frac{17}{2}$ [6+ 96]

$S_{17}$ =  $\frac{17}{2}$  x 102

$S_{17}$ = 17 x 51

$S_{17}$ = 867

Therefore, the sum of all odd integers between 2 and 100 divisible by 3 is 867

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