Advertisements
Advertisements
Question
Find the sum of all integers between 100 and 550, which are divisible by 9.
Advertisements
Solution
In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.
So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.
Also, all these terms will form an A.P. with the common difference of 9.
So here,
First term (a) = 108
Last term (l) = 549
Common difference (d) = 9
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
`a_n = a + (n - 1)d`
So for the last term
`549 = 108 + (n - 1)9
549 = 108 + 9n - 9
549 = 99 + 9n
Further simplifying,
450 = 9n
`n= 450/9`
n = 50
Now, using the formula for the sum of n terms,
`S_n =n/2[2a + (n -1)d]`
We get
`S_n = 50/2 [2(108) + (50 - 1)d]`
= 25[216 + (49)9]
= 25(216 + 441)
= 25(657)
= 16425
Therefore, the sum of all the multiples of 9 lying between 100 and 550 is `S_n= 16425`
APPEARS IN
RELATED QUESTIONS
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
If the term of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m + n) terms is zero
Find the sum of first 15 multiples of 8.
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
Find the sum of the first 11 terms of the A.P : 2, 6, 10, 14, ...
Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
If 4 times the 4th term of an A.P. is equal to 18 times its 18th term, then find its 22nd term.
Find the three numbers in AP whose sum is 15 and product is 80.
Find the sum of first n even natural numbers.
Write an A.P. whose first term is a and common difference is d in the following.
a = –1.25, d = 3
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
Find the sum of n terms of the series \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Suppose the angles of a triangle are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?
Find second and third terms of an A.P. whose first term is – 2 and the common difference is – 2.
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
In an A.P., the sum of first n terms is `n/2 (3n + 5)`. Find the 25th term of the A.P.
