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Hint: calculate the interest at the end of each year using the simple interest formula and then try to generalise the interests obtained in order to show it as a sequence of AP or not. Then using the formula for the nth term of AP series find the interest at the end of 30 years.

Formula for simple interest is

\[\text{SI=}\dfrac{P\times T\times R}{100}............\left( 1 \right)\]

Where SI = simple interest

P= principal

R= rate of interest

T= time

Let us assume that the interests at the end of the years as

\[\text{S}{{\text{I}}_{1}},\text{S}{{\text{I}}_{2}},\text{S}{{\text{I}}_{3}},\text{etc}\]

Given in the question that

P=1000

R=8

Now to calculate the interest at the end of the 1 year we need to take

T=1

By substituting the values of P,R,T in the above simple interest formula (1) we get,

\[\begin{align}

& \text{S}{{\text{I}}_{1}}\text{=}\frac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{1}}=\frac{1000\times 8\times 1}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{1}}=10\times 8 \\

& \therefore \text{S}{{\text{I}}_{1}}=80 \\

\end{align}\]

Now the interest at the end of the 2 year can be calculated similarly by substituting the values

P=1000

R=8

T=2 in the above simple interest formula (1)

\[\begin{align}

& \text{S}{{\text{I}}_{2}}\text{=}\frac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=\frac{1000\times 8\times 2}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=10\times 8\times 2 \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=80\times 2 \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=160 \\

& \therefore \text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\

\end{align}\]

Similarly interest at the end of 3 years can be calculated by substituting the respective values in simple interest formula (1)

P=1000

R=8

T=3

\[\begin{align}

& \text{S}{{\text{I}}_{3}}\text{=}\dfrac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=\dfrac{1000\times 8\times 3}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=10\times 8\times 3 \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=80\times 3 \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=240 \\

& \therefore \text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\

\end{align}\]

Similarly for all other SI also we can write in terms of SI1 [SI1=80]

Now by considering the interests at the end of 1,2,3 years SI1 ,SI2 , SI3 we can observe that

\[\text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}\]

\[\text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}\]

etc

Hence , interest at the end of any year let’s suppose be n can be generalised as

\[\text{S}{{\text{I}}_{n}}=n\times \text{S}{{\text{I}}_{1}}\]

Here we can also observe that

\[\text{S}{{\text{I}}_{2}}-\text{S}{{\text{I}}_{1}}=\text{S}{{\text{I}}_{1}}\]

\[\text{S}{{\text{I}}_{3}}-\text{S}{{\text{I}}_{2}}=\text{S}{{\text{I}}_{1}}\]

etc

\[\text{S}{{\text{I}}_{n}}-\text{S}{{\text{I}}_{n-1}}=\text{S}{{\text{I}}_{1}}\]

As we already know that from the definition,

A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP)

Hence the given series of interests at the end of the years 1,2,3…. forms an AP of common difference SI1 i.e., 80

As we know that in an AP series the \[nth\text{ term}\] can be calculated by using the formula

\[{{\text{a}}_{n}}=a+\left( n-1 \right)d..........(2)\]

Where a= first term

d=common difference

now interest at the end of 30 years can be calculated by using the above formula of AP (2)

\[\begin{align}

& {{\text{a}}_{n}}=a+\left( n-1 \right)d \\

& a=\text{S}{{\text{I}}_{1}}=80 \\

& d=\text{S}{{\text{I}}_{1}}=80 \\

& n=30 \\

\end{align}\]

Now by substituting the values in the above formula by assuming interest at ath end of 30 years as \[\text{S}{{\text{I}}_{30}}\] we get,

\[\begin{align}

& \text{S}{{\text{I}}_{30}}=\text{S}{{\text{I}}_{1}}+\left( n-1 \right)\text{S}{{\text{I}}_{1}} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=80+(30-1)80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=80+29\times 80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=(1+29)\times 80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=30\times 80 \\

& \therefore \text{S}{{\text{I}}_{30}}=2400 \\

\end{align}\]

Hence, the interest at the end of 30 years is Rs.2400

Note:

We need to write the interests at the end of 2,3 years in terms of interest at the end of 1 year so that we can get a generalised form to calculate the interest at the end of any year easily.

Interest at the end of 30 years can also be calculated by using the simple interest formula instead of using the \[nth\text{ term}\] of an AP series formula

\[\begin{align}

& \text{SI=}\dfrac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=\frac{1000\times 30\times 8}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=10\times 30\times 8 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=300\times 8 \\

& \therefore \text{S}{{\text{I}}_{30}}=2400 \\

\end{align}\]

Which gives the same result in either ways.

By writing the generalised form for the interests at the end of each year helps in finding whether the sequence is in AP or not.

Formula for simple interest is

\[\text{SI=}\dfrac{P\times T\times R}{100}............\left( 1 \right)\]

Where SI = simple interest

P= principal

R= rate of interest

T= time

Let us assume that the interests at the end of the years as

\[\text{S}{{\text{I}}_{1}},\text{S}{{\text{I}}_{2}},\text{S}{{\text{I}}_{3}},\text{etc}\]

Given in the question that

P=1000

R=8

Now to calculate the interest at the end of the 1 year we need to take

T=1

By substituting the values of P,R,T in the above simple interest formula (1) we get,

\[\begin{align}

& \text{S}{{\text{I}}_{1}}\text{=}\frac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{1}}=\frac{1000\times 8\times 1}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{1}}=10\times 8 \\

& \therefore \text{S}{{\text{I}}_{1}}=80 \\

\end{align}\]

Now the interest at the end of the 2 year can be calculated similarly by substituting the values

P=1000

R=8

T=2 in the above simple interest formula (1)

\[\begin{align}

& \text{S}{{\text{I}}_{2}}\text{=}\frac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=\frac{1000\times 8\times 2}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=10\times 8\times 2 \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=80\times 2 \\

& \Rightarrow \text{S}{{\text{I}}_{2}}=160 \\

& \therefore \text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\

\end{align}\]

Similarly interest at the end of 3 years can be calculated by substituting the respective values in simple interest formula (1)

P=1000

R=8

T=3

\[\begin{align}

& \text{S}{{\text{I}}_{3}}\text{=}\dfrac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=\dfrac{1000\times 8\times 3}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=10\times 8\times 3 \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=80\times 3 \\

& \Rightarrow \text{S}{{\text{I}}_{3}}=240 \\

& \therefore \text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\

\end{align}\]

Similarly for all other SI also we can write in terms of SI1 [SI1=80]

Now by considering the interests at the end of 1,2,3 years SI1 ,SI2 , SI3 we can observe that

\[\text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}\]

\[\text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}\]

etc

Hence , interest at the end of any year let’s suppose be n can be generalised as

\[\text{S}{{\text{I}}_{n}}=n\times \text{S}{{\text{I}}_{1}}\]

Here we can also observe that

\[\text{S}{{\text{I}}_{2}}-\text{S}{{\text{I}}_{1}}=\text{S}{{\text{I}}_{1}}\]

\[\text{S}{{\text{I}}_{3}}-\text{S}{{\text{I}}_{2}}=\text{S}{{\text{I}}_{1}}\]

etc

\[\text{S}{{\text{I}}_{n}}-\text{S}{{\text{I}}_{n-1}}=\text{S}{{\text{I}}_{1}}\]

As we already know that from the definition,

A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP)

Hence the given series of interests at the end of the years 1,2,3…. forms an AP of common difference SI1 i.e., 80

As we know that in an AP series the \[nth\text{ term}\] can be calculated by using the formula

\[{{\text{a}}_{n}}=a+\left( n-1 \right)d..........(2)\]

Where a= first term

d=common difference

now interest at the end of 30 years can be calculated by using the above formula of AP (2)

\[\begin{align}

& {{\text{a}}_{n}}=a+\left( n-1 \right)d \\

& a=\text{S}{{\text{I}}_{1}}=80 \\

& d=\text{S}{{\text{I}}_{1}}=80 \\

& n=30 \\

\end{align}\]

Now by substituting the values in the above formula by assuming interest at ath end of 30 years as \[\text{S}{{\text{I}}_{30}}\] we get,

\[\begin{align}

& \text{S}{{\text{I}}_{30}}=\text{S}{{\text{I}}_{1}}+\left( n-1 \right)\text{S}{{\text{I}}_{1}} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=80+(30-1)80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=80+29\times 80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=(1+29)\times 80 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=30\times 80 \\

& \therefore \text{S}{{\text{I}}_{30}}=2400 \\

\end{align}\]

Hence, the interest at the end of 30 years is Rs.2400

Note:

We need to write the interests at the end of 2,3 years in terms of interest at the end of 1 year so that we can get a generalised form to calculate the interest at the end of any year easily.

Interest at the end of 30 years can also be calculated by using the simple interest formula instead of using the \[nth\text{ term}\] of an AP series formula

\[\begin{align}

& \text{SI=}\dfrac{P\times T\times R}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=\frac{1000\times 30\times 8}{100} \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=10\times 30\times 8 \\

& \Rightarrow \text{S}{{\text{I}}_{30}}=300\times 8 \\

& \therefore \text{S}{{\text{I}}_{30}}=2400 \\

\end{align}\]

Which gives the same result in either ways.

By writing the generalised form for the interests at the end of each year helps in finding whether the sequence is in AP or not.

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