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#### Justify whether it is true to say that the following are the nth terms of an AP. 1+n+n^2

Solution.

$\text{Here } a_{n} = 1 + n + n^{2} \\ \text{Put } n = 1, \quad a_{1} = 1 + 1 + (1)^{2} = 3 \\ \text{Put } n = 2, \quad a_{2} = 1 + 2 + (2)^{2} = 7 \\ \text{Put } n = 3, \quad a_{3} = 1 + 3 + (3)^{2} = 13 \\ \text{The numbers are } 3, 7, 13, \ldots \\ a_{2} - a_{1} = 7 - 3 = 4 \\ a_{3} - a_{2} = 13 - 7 = 6 \\$

Here common difference is not same

Hence $1 + n + n^2$  is not the nth term of an AP.

#### In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers. The number of bacteria in a certain food item after each second, when they double in every second.

Solution.

Let number of bacteria = x

According to question:

They double in every second.

$\\ \therefore x, 2x, 2(2x), 2(2 \times 2x) \ldots \ldots \ldots \ldots \ldots \\ Here a\textsubscript{1 }= x , a\textsubscript{2 }=2x, a\textsubscript{3 }=4x, a\textsubscript{4}=8x\\ x, 2x, 4x, 8x \ldots .. \ldots ..\\ a\textsubscript{2} - a\textsubscript{1} = 2x - x = x\\ a\textsubscript{3}- a\textsubscript{2} = 4x - 2x = 2x\\$

Since, the difference betweenthe successive terms is not same therefore it does not form an AP.

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#### In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers. The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum

Solution.         We know that : Simple interest = (P × R × T)/100

$\text{Here } P = 1000 \text{ Rs, } R = 10\% \text{ p.a., } T = 1 \text{ year} \\ \text{SI} = \frac{1000 \times 10 \times 1}{100} = 100 \text{ Rs} \\ \text{Here the common difference } (d) = 100 \text{ Rs and the first term } (a_1) = 1000 \text{ Rs} \\ a_1 = 1000 \text{ Rs} \\ a_2 = 1000 + 100 = 1100 \text{ Rs} \\ a_3 = 1000 + 200 = 1200 \text{ Rs} \\ a_4 = 1000 + 300 = 1300 \text{ Rs} \\ \text{So the AP is } 1000, 1100, 1200, 1300, \ldots$

This is definite form an AP with common difference of SI=100 Rs.

#### In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers. The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the

Solution.

According to question:

Fees started Rs. 250 and increased by 50 Rs. from I to XII is

$\text{Here the first term } (a_1) = 250 \text{ and the common difference } (d) = 50 \\ a_1 = 250 \\ a_2 = 250 + 50 = 300 \\ a_3 = 300 + 50 = 350 \\ a_4 = 350 + 50 = 400 \\ \text{So the AP is } 250, 300, 350, 400, \ldots$

It forms an AP with common difference 50.

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#### i)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.ii)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.iii)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.iv)In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.The number of bacteria in a certain food item after each second, when they double in every second.

i)

Solution.

According to the question:

It is clear that the monthly fee charged from students is = 400 Rs.

Hence series is ; 400, 400, 400, 400 ……

the difference of the successive terms is same so It forms an AP with common difference d = 400 – 400 = 0

ii)

Solution.

According to the question:

Fees started Rs. 250 and increased by 50 Rs. from I to XII is

$\\Here first term( a\textsubscript{1})=250 common difference (d)=50 \\ a\textsubscript{1}=250\\ a\textsubscript{2}=250+50\\ =300\\ a\textsubscript{3}=300+50\\ =350\\ a\textsubscript{4}=350+50\\ =400\\ So the A.P. 250, 300, 350, 400 \ldots \\$

It forms an AP with common difference 50.

iii)

Solution.         We know that: Simple interest = (P × R × T)/100

$\text{Here } P = 1000 \text{ Rs, } R = 10\% \text{ p.a., } T = 1 \text{ year} \\ \text{SI} = \frac{1000 \times 10 \times 1}{100} = 100 \text{ Rs} \\ \text{Here common difference } (d) = 100 \text{ Rs and first term } (a_1) = 1000 \text{ Rs} \\ a_1 = 1000 \text{ Rs} \\ a_2 = 1000 + 100 = 1100 \text{ Rs} \\ a_3 = 1000 + 200 = 1200 \text{ Rs} \\ a_4 = 1000 + 300 = 1300 \text{ Rs} \\ \text{So the AP is: } 1000, 1100, 1200, 1300, \ldots$

This is definite form an AP with common difference of SI=100 Rs.

iv)

Solution.

Let the number of bacteria = x

According to the question:

They double in every second.

$\text{Here } a_1 = x, \quad a_2 = 2x, \quad a_3 = 4x, \quad a_4 = 8x, \ldots \\ x, 2x, 4x, 8x, \ldots$

Since the difference between the successive terms is not same therefore it does not form an AP.

#### The taxi fare after each km, when the fare is Rs 15 for the first km and Rs. 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is15, 8, 8, 8, ...Is the statement true? Give reasons.

Solution.

$\text{Given series is } 15, 8, 8, 8, \ldots \\ \text{Here } a_1 = 15, \quad a_2 = 8, \quad a_3 = 8, \quad a_4 = 8 \\ a_2 - a_1 = 8 - 15 = -7 \\ a_3 - a_2 = 8 - 8 = 0 \\ a_4 - a_3 = 8 - 8 = 0$

Here difference of the successive terms is not same hence it does not form an AP

Therefore given statement is true.

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#### Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.

Solution.

Here he given series is 31, 28, 25, …….

$\\a = 31\\ d = a\textsubscript{2}- a\textsubscript{1} = 28 - 31= - 3\\ If 0 is a term of this A.P then value of n must be a positive integer.\\ a\textsubscript{n} = a + (n - 1)d\\ 0 = a + (n - 1)d\\ 0 = 31 + (n - 1) (-3)\\ 0 = 31 - 3n + 3\\ 3n = 34\\ n=\frac{34}{3}=11.333. \\$

Which is not an integer.

Hence 0 is not a term of this AP.

#### Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21stterms, which is the same as the difference between any two corresponding terms. Why?

Solution.

It is given that the first term of an AP is 2 and of other is 7.

$\text{Let } a_1 = 2, \quad b_1 = 7 \\ \text{Common difference of both APs is same.} \\ A_n \text{ is the } n\text{-th term for the first AP, } b_n \text{ is the } n\text{-th term of the second AP.} \\ \text{As per the question:} \\ a_{10} - b_{10} = a_1 + 9d - b_1 - 9d \\ = a_1 - b_1 = 2 - 7 = -5 \\ a_{21} - b_{21} = a_1 + 20d - b_1 - 20d \\ = a_1 - b_1 = -5 \\ \text{Difference between the first terms} = a_1 - b_1 = -5$

it is because when we find the difference between 10th and 21st term it is equal to the a1-b1  which is the difference between first terms.

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#### For the AP: –3, –7, –11, ..., can we find directly $a_{30} - a_{20}$ without actually finding $a_{30} \text{ and } a_{20}$?Give reasons for your answer.

Solution. Here the given series is

$-3,-7,-11 \ldots$
$a_{1}=-3, a_{2}=-7$
\begin{align*} \mathrm{d} &= \mathrm{a}_{2} - \mathrm{a}_{1} \\ &= -7 - 3 \\ &= -7 + 3 \\ &= -4 \\ \mathrm{a}_{30} - \mathrm{a}_{20} &= \mathrm{a} + (30 - 1) \mathrm{d} - (\mathrm{a} + (20 - 1) \mathrm{d}) \\ &= \mathrm{a} + 29 \mathrm{~d} - \mathrm{a} - 19 \mathrm{~d} \\ &= 10 \mathrm{~d} \\ &= 10 \times (-4) \\ &= -40 \end{align*}
$\text{Yes we can find}$ $\mathrm{a}_{30}-\mathrm{a}_{20}$  $\text{without actually calculating}$ $\mathrm{a}_{30} \text{ and } \mathrm{a}_{20}$

#### Justify whether it is true to say that $-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots$  forms an A.P.as  $a_2 - a_1 = a_3 - a_2.$

Solution.

$-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots \ldots \\ \text{here} \ \ \mathrm{a}_{1}=-1 \ \ \mathrm{a}_{2}=-\frac{3}{2} \ \ \mathrm{a}_{3}=-2 \ \ \mathrm{a}_{4}=\frac{5}{2}$
$\\ a_{2}-a_{1}=-\frac{3}{2}-(-1)=\frac{-3}{2}+1=\frac{-3+2}{2}=\frac{-1}{2} \\\\ a_{3}-a_{2}=-2-\left(\frac{-3}{2}\right)=-2+\frac{3}{2}=\frac{-4+3}{2}=\frac{-1}{2}$
$\text{Here}$ $a_{2}-a_{1}=a_{3}-a_{2},$  $\text{Hence it is an A P }$