Progressions - CAT 2011 Sample Questions

Question

A gentleman buys every year Bank's cash certificates of value exceeding the last year's purchase by Rs. 300. After 20 years, he finds that the total value of the certificates purchased by him is Rs. 83,000. Find the value of the certificates purchased by him in the 13^th year.

1. Rs.4900
2. Rs.6900
3. Rs.1300
4. None of these.
   

Correct Choice is (1) and Correct Answer is Rs.4900

Explanatory Answer

Let the value of the certificates purchased in the first year be Rs. a.

The difference between the value of the certificates is Rs.300 (d = 300).

Since, it follows Arithmetic progression the total value of the certificates after 20 years is given by

S_n =  = .

By simplifying, we get 2a + 5700 = 8300.

Therefore, a = Rs.1300.

The value of the certificates purchased by him in nth year = a + (n - 1) d.

Therefore, the value of the certificates purchased by him in 13th year = 1300 + (13 - 1) 300 = Rs.4900.

Question

The sum of the first 50 terms common to the Arithmetic Sequence 15, 19, 23..... and the Arithmetic Sequence 14, 19, 24..... is.

1. 25450
2. 24550
3. 50900
4. 49100
   

Correct Choice is (1) and Correct Answer is 25450

Explanatory Answer

The two series are in A.P. Therefore, the common series will also be in an A.P

Common difference of 1st series = 4 and the common difference of 2nd series = 5.

Common difference of the sequence whose terms are common to the two series is given by L.C.M of 4 and 5 = 20

Here the first term of the identical (common terms) sequence is 19.

We know the sum of first n terms is given by = 

Hence, the sum of first 50 terms of this sequence =  = 25450.

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More Questions on Arithmetic and Geometric Progressions

Question

What is the sum of all positive integers that are multiples of 7 from 200 to 400?

1. 8729
2. 8700
3. 8428
4. 8278
   

Correct Choice is (1) and Correct Answer is 8729

Explanatory Answer

Because 4 is the remainder when we divide 200 by 7 the least number greater than 200 divisible by 7 is 203.
When we divide 400 by 7, we get a remainder of 1.

This implies that the greatest number less than 400, which is divisible by 7 is 399.

Therefore, this is an Arithmetic Progression in which the first term, t₁ = a = 203, the common difference 'd' = 7, and the last term 'l' = 399

Let 'n' be the total number of terms in this series.

The nth term of an Arithmetic Sequence is a_n = a + (n - 1)d

Then 399 = 203 + (n – 1) (7)

Simplifying we get, 7n = 399 – 203 + 7

Or 406 – 203 = 203 or n = 29.

Hence, the required sum is 

Or  = (29) * (301) = 8729

Question

The sum of the first and the 9^th of an arithmetic progression is 24. What is the sum of the first nine terms of the progression?

1. 240
2. 216
3. 108
4. 120
   

Correct Choice is (3) and Correct Answer is 108

Explanatory Answer

The sum of the first 'n' terms of an AP is , where 'a' is the first term and 'l' is the last term of the series.

In this question, we know that the sum of the first and the last term (ninth term) = 24.

Therefore, the sum of first 9 terms =  = 108

Question

The sum of 20 terms of the series -1² + 2² - 3² + 4² - 5² + 6².... is:

1. 210
2. 519
3. 190
4. 420
   

Correct Choice is (1) and Correct Answer is 210

Explanatory Answer

-1² + 2² - 3² + 4² - 5² + 6².... upto 20 terms can be written as

(-1² + 2²) + (-3² + 4²) + (-5² + 6²) + ..... upto 20 terms.

= (-1 + 2) (1 + 2) + (-3 + 4) (3 + 4) + (-5 + 6) (5 + 6) +..... (-19 + 20) (19 + 20)

= (1 + 2) + (3 + 4) + (5 + 6) + ..... + (19 + 20)

Sum of the first 'n' natural numbers = 

Therefore, sum of the given series =  = 210.