How much money is to be invested every year so to accumulate Rs. 3,00,000 at the end of 10 years if interest is compounded annually at 10%
[ A (10,0.1) = 15.9374)
(a) Rs.18823.65
(b) Rs.18833.64
(c) Rs.18223.60
(d) Rs.16823.65
Answers
Step-by-step explanation:
Enclosure consists of solution.
The required investment is 17112.406
Given:
Accumulated amount = Rs. 3,00,000
Time period = 10 years
Rate of compound interest = 10%
To find:
How much money is to be invested every year so as to accumulate Rs. 3,00,000 at the end of 10 years?
Solution:
Let the amount invested is 'P' every year
When the amount 'P' is invested in 1st year then after 10 the amount will be = P(1 + 10/100)¹⁰ = P(1.1)¹⁰
When the amount 'P' is invested in 2nd year then after 9 the amount will be = P(1 + 10/100)⁹ = P(1.1)⁹
If we calculated like this the amounts in 10 years will be as follows
P(1.1)¹⁰ , P(1.1)⁹, P(1.1)⁸ ......P(1.1)¹
Which can be written as
P(1.1)¹, P(1.1)², ... P(1.1)⁸, P(1.1)⁹, P(1.1)¹⁰
As we can see from the series in GP,
Where the first term, a = P(1.1)¹
The common difference, d = (1.1)
Number of terms, n = 10
∵ Sum of n terms in GP = a[(r^n – 1)/(r – 1)]
Sum the amounts from each of the 10 years
= P(1.1)¹[(1.1)¹⁰ – 1)/(1.1 – 1)]
= P(1.1) [ 15.9374 ]
= P(17.53114)
From the given data,
The total amount = 3,00,000
=> P(17.53114) = 300000
=> P = 300000/17.53114
=> P = 17112.406
Therefore,
The required investment is 17112.406
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