Sir/Madam Plss solve it

Sum of all integers between 100 and 500 which are divisible by 17

### Asked by seeni2005 | 17th Aug, 2020, 12:42: AM

Expert Answer:

### The first and last numbers between 100 and 500 which are divisible by 17 are 102 and 493.
The numbers divisible by 17 forms an AP with common difference d=17, first term a = 102 and last term a_{n} = 493
a_{n} = a + (n-1)d
493 = 102 + (n-1)17
(n-1)17 = 391
n-1 = 23
n = 24
Now, the formula for sum of n terms of an AP can be used to find the required sum

_{n}= 493

_{n}= a + (n-1)d

### Answered by Renu Varma | 17th Aug, 2020, 12:19: PM

## Concept Videos

- Arithmetic progression all formulas
- The sum of the first 7 terms of an A.P is 182 if its 4th ans 17th terms are in ratio 1:5 find the Ap
- 0.6,1.7,2.8........to100term
- sn denotes the sum of first n terms of an AP, whose common difference is d, then Sn-2Sn-1 + Sn-2 (n > 2) is equal to A) 2d B) -d C) d D) None of these Explain solution please
- Plz solve this question
- Sum of first m terms of an A.P. is 0. If a be the first term of the A.P., then the sum of next n terms is : (A) 1 ( ) − − + m a m n m (B) 1 ( ) − − + m a m n n (C) 1 ( ) − − + n a m n n (D) 1 ( ) − − + n a m n m
- Question 1) Find the sum of first 40 positive integers divisible by 6. Question 2) Find the sum of n terms of the series (4-1/n)+4-2/n)+(4-3/n)+........ Question 3) If the sum of the first n terms of an AP is 1/2(3n²+7n), then find its nth term .Hence write the 20 th term .
- A3=15,S10=125,find d and a10
- Find the sum of the first 17 term of an AP whose 4th term = -15 and 9th term = -30.

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change