Question

Find the amount of an ordinary annuity of Rs400 payable at the end of every 3 months for 6 years at 8% per annum compounded quarterly. (Use log table)

Answer 1

Given,

R = 400  

r = 0.08/4 = 0.02  

n = 6*4 = 24

Formula

$A=R[\frac{(1+r)^{n}-1}{r}]$

Let      

x = (1.02) ^24

Then,  

Log x = 24 log 1.02

  = 24 * 0.086 = 0.2064

x = antilog 0.2064 = 1.608

Then we have,

A = $400*\frac{1.608-1}{0.02}$

   = $400*\frac{0.608}{0.02}$

 

     = 12160

Hence the amount is Rs 12160.

Answer 2

Answer:

Step-by-step explanation:

We know that,

R=400

r=0.08

-------

4

=0.02

n=6×4

=24

Use the formula,

A=R[(1+r)ⁿ-1]

--------------

r

So,

X=[1.02]*²⁴

Apply log value on both sides ,

Log X= 24 log 1.02

=24×0.086

=0.2064

Where,

X= anti log of 0.2064

X=1.608

Now,

A=400[(1.608-1)]

--------------------

0.02

A=400[(0.608)]

--------------------

0.02

A=243.2

---------

0.02

A=12160

Hope it will help you

✌️Sai