Question

Find the Sum of all integers between 3 and 99 divisible by 3​

Answer 1

Answer :

$\large{\dag\:{\boxed{\bf{\star\:\:S_{33}\:=\:1683\:\:\star}}\:\dag}$

Explanation :

Given :–

• A.P. :- 3 , 6 , 9 , 12 , ... , 99 .
• Where a = 3 , d = 3 and aₙ = 99 .

To Find :–

• Sum of all Integers between 3 to 99 which are divisible by 3 .(Sum of the above A.P.)

Formulas Applied :–

• $\boxed{\bf{\star\:\:a_n\:=\:a\:+\:(n\:-\:1)d\:\:\star}}$

• $\boxed{\bf{\star\:\:S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d] \:\:\star}}$

Solution :–

First we will find the Number of terms (n) in this A.P. :-

We have , a = 3 , d = 3 and aₙ = 99 .

Putting these values in the formula :

$\rightarrow\sf{a_n\:=\:a\:+\:(n\:-\:1)d}$

$\rightarrow\sf{99\:=\:3\:+\:(n\:-\:1)(3)}$

$\rightarrow\sf{99\:=\:3\:+\:3n\:-\:3}$

$\rightarrow\sf{99\:=\:3n}$

$\rightarrow\sf{n\:=\:\dfrac{99}{3} }$

$\rightarrow\bf{n\:=\:33 }$

Now , we have to find S₃₃ :

We have now n = 33 , a = 3 and d = 3 .

Putting these values in the formula :

$\rightarrow\sf{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)d]}$

$\rightarrow\sf{S_{33}\:=\:\dfrac{33}{2}\:[2(3)\:+\:(33\:-\:1)(3)]}$

$\rightarrow\sf{S_{33}\:=\:\dfrac{33}{2}\:[6\:+\:(32\:\times\:3)]}$

$\rightarrow\sf{S_{33}\:=\:\dfrac{33}{2}\:[6\:+\:96]}$

$\rightarrow\sf{S_{33}\:=\:\dfrac{33}{2}\:\times\:102}$

$\rightarrow\sf{S_{33}\:=\:33\:\times\:51}$

$\rightarrow\boxed{\bf{S_{33}\:=\:1683}}$

∴Sum of all integers between 3 and 99 which are divisible by 3 is 1683 .

Answer 2

Answer:

Sum of all integers between 3 and 99 divisible by 3​ = 1581

Step-by-step explanation:

To find,

The sum of all integers between 3 and 99 divisible by 3

Recall the concepts:

The nth term of AP = $a_n$ = a+(n-1)d -------------(1)

The Sum to n terms of an AP = $S_n = \frac{n}{2}[2a +(n-1)d]$ = $\frac{n}{2}[a +l]$,------------(2)

where 'a' is the first term, 'd' is the common difference, 'l' is the last term

Solution:

The numbers divisible by 3 between 3 and 99 form an AP.

The terms of this AP are 6,9,12,......................., 96

here, First term = a = 6

common difference = d = 3

Last term = l = 96

To find 'n'

We have, $a_n$ = a+(n-1)d (from equation (1))

Substitute the values of  $a_n$ = 96, a = 6 and d= 3 we get

96 = 6+(n-1)3

90 = 3(n-1)

30 = n-1

n = 31

∴ There are 31 integers between 3 and 99 which are divisible by 3

Hence we need to find $S_{31}$

From equation(2)

$S_{31} = \frac{31}{2}[a+l]$

Substitute the value of a = 6 and l = 96 we get

$S_{31} = \frac{31}{2}[6+96]$ = $\frac{31}{2} X102$ = 51 × 31 = 1581

∴Sum of all integers between 3 and 99 divisible by 3​ = 1581

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