Answer

Verified

451.5k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE