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Worked-out Problems

Example-1
An investor starts with $ 500 in an investment account, and each month it earns a constant interest of $ 32. After how many months will the sum exceed $ 700 in the account?
  1. 2
  2. 3
  3. 6
  4. 8
Solution
Here, we know that a + (n – 1)d = 700 where a = 500 and d = 32. The reason for not using the sum formula is that the amount in the account, after each successive month, represents each term of the series and not the added up terms.
 
The equation becomes 500 + (n − 1) × 32 = 700. This solution gives n = 7.25, which is rounded up to 8. After eight months the account will hold in excess of $ 700.
 
So, option (d) is the answer.
 
 
Example-2
The sum of an AP, consisting of 12 terms, is 354. The ratio of the sum of the odd terms to the sum of the even terms is 27:32. What is the common difference of this AP?
  1. 2
  2. 3
  3. 4
  4. 5
Solution
Let the common difference be ‘d’ and the terms are T1, T2, T3
So, T2 = T1 + d
And T4 = T3 + d and so on.
So, T2 + T4 + T6 + T8 + T10 + T12 = T1 + T3 + T5 + T7 + T9 + T11 + 6 d …(1)
Sum of the even terms : Sum of the odd terms
= 32:27 and their sum = 354
So, the sum of the even terms = 192 and of the odd terms = 162, Using (1), d = 5
 
 
Example-3
Find the sum of n terms of the series 11 + 103 +1005 +…
  1. 10/9(10n −1) −1
  2. 100/99(10n−1) + n2
  3. (c) 10/9(10n −1) + n2
  4. None of these
Solution
Ideally, in these kinds of problems, instead of going by the mathematical process of solving, we should use options.
 
Checking option (1), Put n = 1.
10/9(10n−1) –1 = 9, so it is not correct.
 
Checking option (2), Put n = 1.
100/99(10n−1) + n2 is not equal to 11, so this is also not correct.
 
Checking option (c), Put n = 1.
10/9(10n−1) + n2 = 11. But just because this option satisfies n = 1, it should not be assumed to be correct. Let us check it for n = 2.
 
Option (c) gives us 104. So, this is the answer.
 
Normally, in these cases, checking the options till n = 2 guarantees the answer, but sometimes we need to check it till n = 3.
 




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