Examples - Chapter 5 Arithmetic Equations class 10 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 5 - Arithmetic Equations - Ncert Solutions class 10 - Maths

Example 1 : .

For the AP : $\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}, \ldots$, write the first term $a$ and the common difference $d$.

Solution :

Here, $a=\frac{3}{2}, d=\frac{1}{2}-\frac{3}{2}=-1$.
Remember that we can find $d$ using any two consecutive terms, once we know that the numbers are in $\mathrm{AP}$.

Example 2:

Which of the following list of numbers form an AP? If they form an AP, write the next two terms:
(i) $4,10,16,22, \ldots$
(ii) $1,-1,-3,-5$,
(iii) $-2,2,-2,2,-2, \ldots$
(iv) $1,1,1,2,2,2,3,3,3, \ldots$

Solution:

(i) We have $a_2-a_1=10-4=6$
$a_3-a_2=16-10=6$
$a_4-a_3=22-16=6$
i.e., $a_{k+1}-a_k$ is the same every time.

So, the given list of numbers forms an AP with the common difference $d=6$.
The next two terms are: $22+6=28$ and $28+6=34$.

(ii)
$
\begin{aligned}
& a_2-a_1=-1-1=-2 \\
& a_3-a_2=-3-(-1)=-3+1=-2 \\
& a_4-a_3=-5-(-3)=-5+3=-2
\end{aligned}
$
i.e., $a_{k+1}-a_k$ is the same every time.

So, the given list of numbers forms an AP with the common difference $d=-2$.
The next two terms are:
$
-5+(-2)=-7 \text { and }-7+(-2)=-9
$

$
\text { (iii) } \begin{aligned}
a_2-a_1 & =2-(-2)=2+2=4 \\
a_3-a_2 & =-2-2=-4
\end{aligned}
$

As $a_2-a_1 \neq a_3-a_2$, the given list of numbers does not form an AP.

(iv)
$
\begin{aligned}
& a_2-a_1=1-1=0 \\
& a_3-a_2=1-1=0 \\
& a_4-a_3=2-1=1
\end{aligned}
$

Here, $a_2-a_1=a_3-a_2 \neq a_4-a_3$.
So, the given list of numbers does not form an AP.

Example 3 :

Find the 10 th term of the AP : $2,7,12, \ldots$
Solution :

Here, $a=2, d=7-2=5$ and $n=10$.
We have
$
\begin{aligned}
& a_n=a+(n-1) d \\
& a_{10}=2+(10-1) \times 5=2+45=47
\end{aligned}
$

So,
Therefore, the 10th term of the given AP is 47 .
Example 4 : Which term of the AP : $21,18,15, \ldots$ is -81 ? Also, is any term 0 ? Give reason for your answer.
Solution : Here, $a=21, d=18-21=-3$ and $a_n=-81$, and we have to find $n$.
As
$
a_n=a+(n-1) d \text {, }
$
we have
$
\begin{aligned}
-81 & =21+(n-1)(-3) \\
-81 & =24-3 n \\
-105 & =-3 n
\end{aligned}
$

So,
$
n=35
$

Therefore, the 35th term of the given AP is -81 .

Next, we want to know if there is any $n$ for which $a_n=0$. If such an $n$ is there, then
$
21+(n-1)(-3)=0,
$
i.e.,
$
\begin{aligned}
3(n-1) & =21 \\
n & =8
\end{aligned}
$
i.e.,

So, the eighth term is 0 .

Example 5 :

Determine the AP whose 3rd term is 5 and the 7 th term is 9 .
Solution :

We have
and
$
\begin{aligned}
& a_3=a+(3-1) d=a+2 d=5 \\
& a_7=a+(7-1) d=a+6 d=9
\end{aligned}
$

Solving the pair of linear equations (1) and (2), we get
$
a=3, \quad d=1
$

Hence, the required $A P$ is $3,4,5,6,7, \ldots$

Example 6 :

Check whether 301 is a term of the list of numbers $5,11,17,23, \ldots$
Solution :

We have :
$
a_2-a_1=11-5=6, \quad a_3-a_2=17-11=6, \quad a_4-a_3=23-17=6
$

As $a_{k+1}-a_k$ is the same for $k=1,2,3$, etc., the given list of numbers is an AP.
Now, $\quad a=5$ and $d=6$.
Let 301 be a term, say, the $n$th term of this AP.
We know that

So,
$
\begin{aligned}
a_n & =a+(n-1) d \\
301 & =5+(n-1) \times 6 \\
301 & =6 n-1
\end{aligned}
$

So,
$
n=\frac{302}{6}=\frac{151}{3}
$

But $n$ should be a positive integer (Why?). So, 301 is not a term of the given list of numbers.

Example 7 :

How many two-digit numbers are divisible by 3 ?
Solution :

The list of two-digit numbers divisible by 3 is :
$
12,15,18, \ldots, 99
$

Is this an AP? Yes it is. Here, $a=12, d=3, a_n=99$.
As
$
\begin{aligned}
& a_n=a+(n-1) d, \\
& 99=12+(n-1) \times 3 \\
& 87=(n-1) \times 3
\end{aligned}
$

i.e.,
$
87=(n-1) \times 3
$
i.e.,
$
\begin{aligned}
n-1 & =\frac{87}{3}=29 \\
n & =29+1=30
\end{aligned}
$

So, there are 30 two-digit numbers divisible by 3 .

Example 8 :

Find the 11th term from the last term (towards the first term) of the AP : $10,7,4, \ldots,-62$.
Solution :

Here, $a=10, d=7-10=-3, l=-62$,
where
$
l=a+(n-1) d
$

To find the 11th term from the last term, we will find the total number of terms in the AP.
So,
$
\begin{aligned}
& -62=10+(n-1)(-3) \\
& -72=(n-1)(-3)
\end{aligned}
$

i.e.,
$
\begin{aligned}
n-1 & =24 \\
n & =25
\end{aligned}
$
or
So, there are 25 terms in the given AP.
The 11th term from the last term will be the 15 th term. (Note that it will not be the 14th term. Why?)
So,
$
a_{15}=10+(15-1)(-3)=10-42=-32
$
i.e., the 11 th term from the last term is -32 .

Alternative Solution :
If we write the given AP in the reverse order, then $a=-62$ and $d=3$ (Why?)

So, the question now becomes finding the 11 th term with these $a$ and $d$.
So,
$
a_{11}=-62+(11-1) \times 3=-62+30=-32
$

So, the 11 th term, which is now the required term, is -32 .

Example 9: A sum of Rs.1000 is invested at $8 \%$ simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.
Solution : We know that the formula to calculate simple interest is given by
$
\text { Simple Interest }=\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}
$

So, the interest at the end of the 1 st year $=Rs.\frac{1000 \times 8 \times 1}{100}=Rs. 80$
The interest at the end of the 2nd year $=Rs. \frac{1000 \times 8 \times 2}{100}=Rs.160$
The interest at the end of the 3rd year $=Rs.\frac{1000 \times 8 \times 3}{100}=Rs. 240$
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.
So, the interest (in Rs.) at the end of the 1st, 2nd, 3rd, ... years, respectively are
$
80,160,240, \ldots
$

It is an AP as the difference between the consecutive terms in the list is 80 , i.e., $d=80$. Also, $a=80$.

So, to find the interest at the end of 30 years, we shall find $a_{30}$.
Now,
$
a_{30}=a+(30-1) d=80+29 \times 80=2400
$

So, the interest at the end of 30 years will be $Rs.2400$.
Example 10 :

In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Solution :

The number of rose plants in the 1 st, 2 nd, $3 \mathrm{rd}, \ldots$, rows are :
$
23,21,19, \ldots, 5
$

It forms an AP (Why?). Let the number of rows in the flower bed be $n$.
Then
$
a=23, \quad d=21-23=-2, a_n=5
$

As,
$
a_n=a+(n-1) d
$

We have,
$
5=23+(n-1)(-2)
$
i.e.,
$
-18=(n-1)(-2)
$
i.e.,
$
n=10
$

So, there are 10 rows in the flower bed.

Example 11 :

Find the sum of the first 22 terms of the AP : $8,3,-2, \ldots$
Solution :

Here, $a=8, d=3-8=-5, n=22$.
We know that
$
\begin{aligned}
& \mathrm{S}=\frac{n}{2}[2 a+(n-1) d] \\
& \mathrm{S}=\frac{22}{2}[16+21(-5)]=11(16-105)=11(-89)=-979
\end{aligned}
$

So, the sum of the first 22 terms of the AP is -979 .

Example 12 :

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution :

Here, $\mathrm{S}_{14}=1050, n=14, a=10$.
As
$
\mathrm{S}_n=\frac{n}{2}[2 a+(n-1) d],
$
so,
$
1050=\frac{14}{2}[20+13 d]=140+91 d
$

i.e.,
$
910=91 d
$
or,
$
d=10
$

Therefore,
$
a_{20}=10+(20-1) \times 10=200 \text {, i.e. 20th term is } 200 \text {. }
$

Example 13 :

How many terms of the AP : $24,21,18, \ldots$ must be taken so that their sum is 78 ?

Solution :

Here, $a=24, d=21-24=-3, \mathrm{~S}_n=78$. We need to find $n$.
We know that
$
\begin{aligned}
& \mathrm{S}_n=\frac{n}{2}[2 a+(n-1) d] \\
& 78=\frac{n}{2}[48+(n-1)(-3)]=\frac{n}{2}[51-3 n]
\end{aligned}
$

So,
$
3 n^2-51 n+156=0
$
or
$
n^2-17 n+52=0
$
or
$
(n-4)(n-13)=0
$
or
$
n=4 \text { or } 13
$
Both values of $n$ are admissible. So, the number of terms is either 4 or 13 .
Remarks:
1. In this case, the sum of the first 4 terms $=$ the sum of the first 13 terms $=78$.
2. Two answers are possible because the sum of the terms from 5 th to 13 th will be zero. This is because $a$ is positive and $d$ is negative, so that some terms will be positive and some others negative, and will cancel out each other.

Example 14 :

Find the sum of :
(i) the first 1000 positive integers
(ii) the first $n$ positive integers

Solution :
(i) Let $\mathrm{S}=1+2+3+\ldots+1000$
Using the formula $\mathrm{S}_n=\frac{n}{2}(a+l)$ for the sum of the first $n$ terms of an $\mathrm{AP}$, we have
$
\mathrm{S}_{1000}=\frac{1000}{2}(1+1000)=500 \times 1001=500500
$

So, the sum of the first 1000 positive integers is 500500 .
(ii) Let $\mathrm{S}_n=1+2+3+\ldots+n$
Here $a=1$ and the last term $l$ is $n$.

Therefore,
$
\mathrm{S}_n=\frac{n(1+n)}{2} \text { or } \mathrm{S}_n=\frac{n(n+1)}{2}
$

So, the sum of first $\boldsymbol{n}$ positive integers is given by
$
\mathrm{S}_n=\frac{n(n+1)}{2}
$

Example 15 :

Find the sum of first 24 terms of the list of numbers whose $n$th term is given by
$
a_n=3+2 n
$

Solution :
As
$
\begin{aligned}
& a_n=3+2 n, \\
& a_1=3+2=5 \\
& a_2=3+2 \times 2=7 \\
& a_3=3+2 \times 3=9
\end{aligned}
$
so,
:

List of numbers becomes $5,7,9,11, \ldots$
Here,
$
7-5=9-7=11-9=2 \text { and so on. }
$

So, it forms an AP with common difference $d=2$.
To find $\mathrm{S}_{24}$, we have $n=24, \quad a=5, \quad d=2$.
Therefore,
$
\mathrm{S}_{24}=\frac{24}{2}[2 \times 5+(24-1) \times 2]=12[10+46]=672
$

So, sum of first 24 terms of the list of numbers is 672 .

Example 16 :

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the 1st year
(ii) the production in the 10th year
(iii) the total production in first 7 years

Solution :

(i) Since the production increases uniformly by a fixed number every year, the number of TV sets manufactured in 1st, 2nd, 3rd, .... years will form an AP.
Let us denote the number of TV sets manufactured in the $n$th year by $a_n$. Then,
$
a_3=600 \text { and } a_7=700
$

or,
$
a+2 d=600
$
and
$
a+6 d=700
$

Solving these equations, we get $d=25$ and $a=550$.
Therefore, production of TV sets in the first year is 550 .
(ii) Now
$
a_{10}=a+9 d=550+9 \times 25=775
$

So, production of TV sets in the 10th year is 775 .
(iii) Also,
$
\begin{aligned}
\mathrm{S}_7 & =\frac{7}{2}[2 \times 550+(7-1) \times 25] \\
& =\frac{7}{2}[1100+150]=4375
\end{aligned}
$

Thus, the total production of TV sets in first 7 years is 4375 .