Question

Taslimuddinahmed60 please answer this I will Mark you brainlist​

Answer 1

$\large\underline{\sf{Solution-}}$

Principal, P = Rs 256

Rate of interest, r = 100 % per annum compounded quarterly.

Time, n = 1 year

We know,

Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is

$\boxed{ \bf{ \: \tt{Amount \: = \: P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}}}$

So, on substituting all the values, we get

$\rm :\longmapsto\:\tt{Amount \: = \: 256\bigg(1+\dfrac{100}{400}\bigg)^{4 \times 1}}$

$\rm :\longmapsto\:\tt{Amount \: = \: 256\bigg(1+\dfrac{1}{4}\bigg)^{4}}$

$\rm :\longmapsto\:\tt{Amount \: = \: 256\bigg(\dfrac{4 + 1}{4}\bigg)^{4}}$

$\rm :\longmapsto\:\tt{Amount \: = \: 256\bigg(\dfrac{5}{4}\bigg)^{4}}$

$\rm :\longmapsto\:\tt{Amount \: = \: 256 \times \bigg(\dfrac{625}{256}\bigg)^{}}$

$\bf\implies \:Amount = 625$

Now, We know that,

Compound interest on a certain sum of money of Rs P invested at the rate of r % per annum compounded quarterly for n years is

$\boxed{ \bf{ \: \tt{ CI=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}-P}}}$

or

$\boxed{ \bf{ \: Compound \: Interest \: = \: Amount - Principal}}$

So,

We have

Principal = Rs 256

Amount = Rs 625

Thus,

$\rm :\longmapsto\: \: Compound \: Interest \: = \: Amount - Principal$

$\rm :\longmapsto\: \: Compound \: Interest \: = 625 - 256$

$\bf :\longmapsto\: \: Compound \: Interest \: = 369$

Hence,

• Compound Interest = Rs 369

Additional Information :-

1. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded annually for n years is

$\boxed{ \bf{ \: \tt{Amount \: = \: P\bigg(1+\dfrac{r}{100}\bigg)^{n}}}}$

2. Amount on a certain sum of money of Rs P invested at the rate of r % per annum compounded semi - annually for n years is

$\boxed{ \bf{ \: \tt{Amount \: = \: P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}}}$

Answer 2

Answer:

Appropriate Cuestion

In an APA, d, prove that

Solution

Let assume that

First term of an AP Common difference of an AP d

WE KNOW THAT,

term of on aithmetic progression

is the tem

a is the first term of the progression

is the no. of term dis the common cifference

THus, According to statement

And

(n-1)= m

On Subtracting equation (2)

esquation, we get

im-nid (m-n

On substituting the value or d nequit.on

<= n = 1+m-1

Now. Consider

On substituting the values of a anc d, we

get

Hanco,

P

Additional Information

-Sum of n terms of an arthmetic

progression

WHERE.

S the sum of n terms of AP

a is the first torm of the progression

15 the no. of term

dis the con cifference