Math, asked by Berseria, 23 days ago

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

Answered by shiwkishor
8

Step-by-step explanation:

Enclosure consists of solution.

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Answered by Dhruv4886
0

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  

#SPJ3

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