Series

What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?

1. 897
2. 164,850
3. 164,749
4. 149,700
5. 156,720

How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

1. 128
2. 142
3. 143
4. 141
5. 129

The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with (m + 2)?

1. m + 4
2. n + 6
3. n + 3
4. m + 5
5. n + 4

The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

1. 300
2. 120
3. 150
4. 170
5. 270

If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?

1. 3
2. 1/3
3. 2
4. 9
5. 1/9

Answers:

1The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.

The next number that will leave a remainder of 2 when divided by 3 is 104, 107, .... 

The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.

So, the given series is an AP with the first term being 101 and the last term being 998 and thhe common difference being 3.

Sum of an AP =  


We know that in an A.P., the nth term a_n = a₁ + (n - 1)*d

In this case, therefore, 998 = 101 + (n - 1)* 3

i.e., 897 = (n - 1) * 3

Therefore, n - 1 = 299 

Or n = 300.

Sum of the AP will therefore, be  = 164,850

2.The smallest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 103.

The largest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 999.

The series of numbers that satisfy the condition that the number should leave a remainder of 5 when divided by 7 is an A.P (arithmetic progression) with the first term being 103 and the last term being 999 having a common difference of 7.

We know that in an A.P, 'l' the last term is given by l = a + (n - 1) * d, where 'a' is the first term, 'n' is the number of terms of the series and 'd' is the common difference.

Therefore, 999 = 103 + (n - 1) * 7

Or 999 - 103 = (n - 1) * 7
Or 896 = (n - 1) * 7
Or n - 1 = 128
Or n = 129

3.The fastest way to solve problems of this kind is to take numerical examples.

The average of 5 consecutive integers from 1 to 5 is 3. Therefore, the value of m is 1 and the value of n is 3.

Now, the average of 9 consecutive integers starting from m + 2 will be average of integers from 3 to 11.

The average of numbers from 3 to 11 is 7.

Now look at the answer choices. Only choice (E) satisfies this condition.

4.The sum of the first n terms of a G.P. is given by , where 'a' is the first term of the G.P., 'r' is the common ratio and 'n' is the number of terms in the G.P.

Therefore, the sum of the first 6 terms of the G.P will be equal to 

And sum of the first 3 terms of the G.P. will be equal to  

The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1 

i.e.  

=>  

=> r³ + 1 = 9 

=> r³ = 8 

=> r = 2