Advertisements
Advertisements
Question
Rs 1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.
Advertisements
Solution
It is given that,
Principal (P) = Rs 1000
Rate (R) = 10%
Simple interest (S.I.) =\[\frac{P \times R \times T}{100}\]
Simple interest after an year = \[\frac{1000 \times 10 \times 1}{100} = Rs 100\]
Simple interest after 2 years = \[\frac{1000 \times 10 \times 2}{100} = Rs 200\]
Simple interest after 3 years = \[\frac{1000 \times 10 \times 3}{100} = Rs 300\]
Hence, the total interest amount is in A.P. i.e. 100, 200, 300,....
Here,
a = 100
d = 100
Now,
\[a_n = a + \left( n - 1 \right)d\]
\[ a_{20} = a + \left( 20 - 1 \right)d\]
\[ = 100 + 19\left( 100 \right)\]
\[ = 100 + 1900\]
\[ = 2000\]
Hence, the interest amount after 20 years is Rs 2000.
APPEARS IN
RELATED QUESTIONS
How many terms of the A.P. 65, 60, 55, .... be taken so that their sum is zero?
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Find the sum 3 + 11 + 19 + ... + 803
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Which term of the AP 21, 18, 15, …… is -81?
If 4 times the 4th term of an A.P. is equal to 18 times its 18th term, then find its 22nd term.
Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
How many terms of the AP 21, 18, 15, … must be added to get the sum 0?
Find the first term and common difference for the A.P.
127, 135, 143, 151,...
In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.
The sum of first n odd natural numbers is ______.
If 18, a, b, −3 are in A.P., the a + b =
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
Find the sum of all the 11 terms of an A.P. whose middle most term is 30.
