Learn Series and Sequences concept with problems for CAT Quantitative Aptitude

You will understand this article more if you have gone through the Article on AP, GP and HP. Let us begin by reviewing some solved application based problems.

Example:

At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was Rs 1250, find the value at the end of 5 years.

Solution:

After each year the value of the machine is 80% of its value the previous year (as it depreciates 20%) so at the end of 5 years the machine will depreciate by 5 times.

Hence, we have to find the 6^th term of the G.P. whose first term a₁ is 1250 and common ratio r is 0.8.Hence, the value at the end 5 years = T₆ = a₁r⁵ = 12.50 (0.8)⁵ = 409.6

Example:

The number of terms of the series 26,21,16,11... to be added so as to get the sum 74 is:

[1] 3

[2] 4

[3] 5

[4] 6

Solution:

In the given series: a= 26 and d = -5

Suppose sum of the first n terms is 74. This implies that —

Example:

The sums of the first ‘n’ terms of two arithmetic series are in the ratio(7_n+1): (4_n+27) . The ratio of their 11^th terms is:

[1]  2:3

[2]  3:2

[3]  3:4

[4]  4:3

Solution:

It is given that  

Therefore, using If T_n and T'_n are n^th the  terms of two arithmetic progression and S_n and S_'n are their sums of the first n terms, respectively then,

Substituting n = 11 in this equation we get

Example:

If the numbers 3^2a-1, 14, 3^4-2a (0<a<1) are the first three terms of an AP, then its fifth term is equal to

[1] 33

[2] 43

[3] 53

[4] 63

Solution:
The first three terms of the AP are: 3^2a-1, 14, 3^4-2a

Therefore, using a + c = 2b, where a, b, c are in AP, we have:

Substituting 9^a = x , we get

        
 
This gives

9^a = 81 or 9^a = 3

Therefore, the numbers are 1,14,27 which are in AP with common difference 13.

Therefore, the fifth term is 1 + 4 x 13 = 53.