CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)

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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)
Exercise 5.2

CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)

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1. Fill in the blanks in the following table, given that $ a_{n} $ is the first term, d the common difference and the nth term of the AP:
$\begin{array}{|c|c|c|c|c|}\hline & a & d & n & a_{n} \\\hline (i) & 7 & 3 & 8 & .... \\\hline (ii) & -18 & .... & 10 & 0 \\\hline (iii) & .... & -3 & .... & -5 \\\hline (iv) & -18.9 & 2.5 & 105 & 3.6 \\\hline (v) & 3.5 & 0 & & .... \\\hline\end{array}$

Answer

(i) Given: $ \mathrm{a}=7, \mathrm{d}=3 $ and $ \mathrm{n}=8 $,
$ a_{n}=\text { ? } $
We know:
$ a_{n}=a+(n-1) d $
Thus, $ \mathrm{a}_{\mathrm{n}}=7+(8-1) 3=7+21=28 $
(ii) Given $ a=-18, n=10, a_{n}=0, d= $ ?
We know that $ \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
Thus, $ 0=-18+(10-1) \mathrm{d} $
$ 0=-18+9 d $
Or, $ 9 \mathrm{d}=18 $
Or, $ d=\frac{18 }{ 9}=2 $
(iii) Given $ d=-3, n=18, a_{n}=-5, a= $ ?
We know that, $ a_{n}=a+(n-1) d $
Or, $ -5=\mathrm{a}+(18-1)(-3) $
Or, $ -5=\mathrm{a}-51 $
Or, $ \mathrm{a}=-5+51=46 $
(iv) Given $ \mathrm{a}=-18.9, \mathrm{d}=2.5, \mathrm{a}_{\mathrm{n}}=3.6, \mathrm{n}= $ ?
We know that, $ a_{n}=a+(n-1) d $
Or, $ 3.6=-18.9+(\mathrm{n}-1) 2.5 $
Or, $ 2.5(\mathrm{n}-1)=3.6+18.9=22.5 $
$ \mathrm{n}-1=\frac{22.5 }{ 2.5}=9 $
$ \mathrm{n}=9+1=10 $
(v) Given that $ \mathrm{a}=3.5, \mathrm{~d}=0, \mathrm{n}=105, \mathrm{a}_{\mathrm{n}}= $ ?
We know:,
$ \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$\mathrm{a}_{\mathrm{n}}=3.5+(105-1) 0$
$\mathrm{a}_{\mathrm{n}}=3.5+0$
$=3.5$

2. Choose the correct choice in the following and justify:

(i) 30th term of the AP: $ 10,7,4, \ldots $, is
A. 97 B. 77 C. $-77$ D. $-87$

Answer

30th term of the AP: $ 10,7,4, \ldots $, is
Here, $ \mathrm{a}=10, \mathrm{~d}=-3 $ and $ \mathrm{n}=30 $
We know:
$ a_{n}=a+(n-1) d$
$\text {or, } a_{30}=10+(30-1)(-3)$
$=10+29(-3)$
$=10-87$
$=-77$
Hence,
Answer (C) $- 77$

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(ii) 11th term of the AP: $ -3,-\frac{1}{2}, 2, \ldots $, is
A. 28 B. 22 C. $-38$ D. $-48$

Answer

11th term of the AP: $ -3,-\frac{1}{2}, 2, \ldots $, is
Here, $ a=-3, d=\frac{5 }{ 2 }$ and $ n=11 $
We know that, $ a_{n}=a+(n-1) d $
$\text { Or, } a_{11}=-3+(11-1) \frac{5 }{ 2}$
$=-3+10 \times(\frac{5 }{ 2})$
$=-3+(5)(5)=-3+25=22$
Hence,
Answer (B) 22

3. In the following APs, find the missing terms in the boxes :

(i) $ 2, \square, 26 $

Answer

(ii) $ \square, 13, \square, 3 $

Answer

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
Let three terms $ \mathrm{a}, \mathrm{b} $ and $\mathrm{c}$ are in A.P Therefore, $ \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b} $ Rearranging we get,
$ b=\frac{a+c}{2} $
So, the middle term is the average of two numbers of an A.P And similarly,
$ \mathrm{a}=2 \mathrm{~b}-\mathrm{c} $ and $ c=2 b-a $
And also nth term of an AP is given by $ a_{n}=a+(n-1) d $
where, $ a= $ first term $ \mathrm{n}= $ number of terms $ \mathrm{d}= $ common difference
We know: In AP, middle term is average of the other two terms
The middle term between 13 and 3 will be;
$\frac{ (13+3) }{ 2}=\frac{16 }{ 2}=8 $
Now, $ \mathrm{a}_{4}-\mathrm{a}_{3}=3-8=-5 $
$ a_{3}-a_{2}=8-13=-5 $
Thus, $ \mathrm{a}_{2}-\mathrm{a}_{1}=-5 $
Or, $ 13-\mathrm{a}_{1}=-5 $
Or, $ \mathrm{a}_{1}=13+5=18 $
Thus, above AP can be written as $ 18,13,8,3 $.

(iii) $ 5, \square, \square, 9\frac{1 }{ 2} $

Answer

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
Let three terms $ \mathrm{a}, \mathrm{b} $ and $\mathrm{c}$ are in A.P Therefore, $ \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b} $ Rearranging we get,
$ b=\frac{a+c}{2} $
So, the middle term is the average of two numbers of an A.P And similarly,
$ \mathrm{a}=2 \mathrm{~b}-\mathrm{c} $ and $ c=2 b-a $
And also nth term of an AP is given by $ a_{n}=a+(n-1) d $
where, $ a= $ first term $ \mathrm{n}= $ number of terms $ \mathrm{d}= $ common difference
We have, $ \mathrm{a}=5 $ and $ \mathrm{a}_{4}=9 \frac{1}{2}=\frac{19}{2} $
We know, nth term of an AP is
$ a_{n}=a+(n-1) d $
where a and d are first term and common difference respectively.
Now common difference:
$\mathrm{a}_{4}=\mathrm{a}+3 \mathrm{d}$
$\frac{19}{2}=5+3 \mathrm{d}$
$3 d=\frac{19}{2}-5$
$3 d=\frac{9}{2}$
$ d=\frac{3}{2} $
Hence, using d, $ 2^{\text {nd }} $ term and $ 3^{\text {rd }} $ term can be calculated as:
$ a_{2}=a+d$
$=5+\frac{3}{2}$
$=\frac{13}{2}$
$=6 \frac{1}{2}$
$a_{3}=a+2 d$
$=5+2 \times \frac{3}{2}$
$=8$
Therefore, the A.P. can be written as: $ 5,6 \frac{1}{2}, 8,9 \frac{1}{2} $

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(iv) $ -4, \square, \square, \square, \square, 6 $

Answer

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
Let three terms $ \mathrm{a}, \mathrm{b} $ and $\mathrm{c}$ are in A.P Therefore, $ \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b} $ Rearranging we get,
$ b=\frac{a+c}{2} $
So, the middle term is the average of two numbers of an A.P And similarly,
$ \mathrm{a}=2 \mathrm{~b}-\mathrm{c} $ and $ c=2 b-a $
And also nth term of an AP is given by $ a_{n}=a+(n-1) d $
where, $ a= $ first term $ \mathrm{n}= $ number of terms $ \mathrm{d}= $ common difference
Here, $ a=-4 $ and $ \mathrm{a}_{6}=6 $
We know, nth term of an AP is
$ a_{n}=a+(n-1) d $
where $ a $ and $ d $ are first term and common difference respectively.
Common difference:
$a_{6}=a+5 d$
$6=-4+5 d$
$5 d=6+4=10$
$d=2$
The second, third, fourth and fifth terms of this AP are:
$a_{2}=a+d=-4+2=-2$
$a_{3}=a+2 d=-4+4=0$
$a_{4}=a+3 d=-4+6=2$
$a_{5}=a+4 d=-4+8=4$
Thus, the given AP can be written as: $ -4,-2,0,2,4,6 $

(v) $ \square, 38, \square, \square, \square,-22 $

Answer

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.
Let three terms $ \mathrm{a}, \mathrm{b} $ and $\mathrm{c}$ are in A.P Therefore, $ \mathrm{b}-\mathrm{a}=\mathrm{c}-\mathrm{b} $ Rearranging we get,
$ b=\frac{a+c}{2} $
So, the middle term is the average of two numbers of an A.P And similarly,
$ \mathrm{a}=2 \mathrm{~b}-\mathrm{c} $ and $ c=2 b-a $
And also nth term of an AP is given by $ a_{n}=a+(n-1) d $
where, $ a= $ first term $ \mathrm{n}= $ number of terms $ \mathrm{d}= $ common difference
Given: Second term $ =38 $ and sixth term $ =-22 $
So, $ a+d=38 \quad\ldots \ldots \text{(1)} $
a $ =38-\mathrm{d}$
$a+5 \mathrm{d}=-22\quad \ldots \ldots\text{(2)} $
Putting the value of a in equation 2, we get,
$ 38 -\mathrm{d}+5 \mathrm{d}=-22$
$38+4 \mathrm{d}=-22$
$4 \mathrm{d}=-22-38$
$4 \mathrm{d}=-60$
$ \mathrm{d}=-15 $
Putting the value of $ d $ in equation 1,
we get, $ a-15=38 $
$\mathrm{a}=38+15=53 $
Therefore, the series is $ 53,38,23,8,-7,-22 $.

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4. Which term of the $ \mathrm{AP}: 3,8,13,18, \ldots $, is 78 ?

Answer

Given,
First term, $ a=3 $, Common difference, $ d=a_{2}-a_{1}=8-3=5 ,$
nth term, $ \mathrm{a}_{\mathrm{n}}=78 $
To find: $ \mathrm{n}= $ ?
We know that nth term of an A.P is given by:
$\mathrm{an}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$78=3+(\mathrm{n}-1) 5$
$(\mathrm{n}-1) 5=78-3=75$
$5 \mathrm{n}-5=75$
$5 \mathrm{n}=80 \mathrm{n}=\frac{80 }{ 5\mathrm{n}} =16$
Thus, 78 is the 16th term of given AP.

5. Find the number of terms in each of the following APs :

(i) $ 7,13,19, \ldots, 205 $

Answer

To find: Number of terms, $ n $
Given: $ \mathrm{a}=7, \mathrm{d}=6, \mathrm{a}_{\mathrm{n}}=205, \mathrm{n}= $ ?
We know that nth term of an AP is given by, $ a_{n}=a+(n-1) d $
$205=7+(n-1) 6$
$(n-1) 6=205-7=198$
$n-1=\frac{198}{6}$
$n-1=33$
$ \mathrm{n}=34 $
Thus, 205 is the 34th of term of this AP.

(ii) $ 18,15 \frac{1}{2}, 13, \ldots,-47 $

Answer

To find: Number of terms, $ n $
Given: $ \mathrm{a}=18, \mathrm{d}=15.5-18=-2.5, \mathrm{a}_{\mathrm{n}}=-47 $
We know that nth term of an AP is given by, $ a_{n}=a+(n-1) d $
Or, $ -47=18+(n-1)(-2.5) $
Or, $ (\mathrm{n}-1)(-2.5)=-47-18=-65 $
$ n-1=\frac{-65}{-2.5} $
Or, $ \mathrm{n}-1=26 $
Or, $ \mathrm{n}=27 $
Thus, -47 is the 27 th term of this AP.

6. Check whether $-150$ is a term of the AP : $ 11,8,5,2 \ldots $

Answer

To find: Whether $-150$ is term of the AP
Given: $ \mathrm{a}=11, \mathrm{d}=8-11=-3,
\mathrm{a}_{\mathrm{n}}=-150, \mathrm{n}= $ ?
We know that $ a_{n}=a+(n-1) d $
where $ a_{n}= $ nth term of AP
$ \mathrm{a}= $ first terms of AP
$ \mathrm{n}= $ no. of terms of AP
$ d= $ common difference of AP Applying the formula, we get
Or, $ -150=11+(\mathrm{n}-1)(-3) $
Or, $ (n-1)(-3)=-150-11=-161 $
Or, $ \mathrm{n}-1=\frac{161 }{ 3} $
$ \mathrm{n}-1=53.67 $
It is clear that 161 is not divisible by three and we shall get a fraction as a result. But number of term cannot be a fraction.
Hence, $- 150$ is not a term of the given AP.

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73

Answer

To Find : 31st term.
Given: 11th term of $ \mathrm{AP}, \mathrm{a}_{11}=38 $ and 16th term of $ \mathrm{AP}, \mathrm{a}_{16}=73 $
We know that $ a_{n}=a+(n-1) d $
where, $ a_{n}= $ nth terms of AP
$ \mathrm{a}= $ first term of $ \mathrm{APn}= $ number of terms $ \mathrm{d}= $ common difference
Hence,
$ \mathrm{a}_{11}=\mathrm{a}+(11-1) \mathrm{d} $
$ a_{11}=a+10 d=38\quad \ldots\ldots\text{(i)} $
And,
$a_{16}=a+(16-1) d$
$a_{16}=a+15 d=73\quad \ldots\ldots\text{(ii)}$
Subtracting eq(i) from eq(ii), we get following:
$ a+15 d-(a+10 d)=73-38 $
$ a+15 d-a-10 d=35 $
Or, $ 5 \mathrm{d}=35 $
Or, $ \mathrm{d}=7 $
Substituting the value of $ d $ in eq(i) we get;
$ a+10 \times 7=38 $
Or, $ a+70=38 $
Or, $ \mathrm{a}=38-70=-32 $
Now 31st term can be calculated as follows:
$ a_{31}=a+(31-1) d $
$a_{31}=a+30 d$
$=-32+30 \times 7$
$=-32+210=178$
So, 31st term is 178.

8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the $ 29^{\text {th}} $ term.

Answer

Given, $ \mathrm{a}_{3}=12 $ and $ \mathrm{a}_{50}=106 $
We know that nth term of an AP is given by
$ a_{n}=a+(n-1) d $
where, $ a_{n}= $ nth term of the AP
$ \mathrm{a}= $ first term of APn = number of terms of APd = common difference of AP. Applying the formula for 3rd and 50th term we get,
$\mathrm{a}_{3}=\mathrm{a}+2 \mathrm{d}=12$
$\mathrm{a}_{50}=\mathrm{a}+49 \mathrm{d}=106$
Subtracting 3rd term from 50th term, we get;
$a+49 d-a-2 d=106-12$
$47 d=94$
$d=2$
Substituting the value of $ d $ in 3rd term, we get;
$ a+2 \times 2=12 $
Or, $ a+4=12 $
Or, $ \mathrm{a}=8 $
Now, 29th term can be calculated as follows:
$\mathrm{a}_{29}=\mathrm{a}+28 \mathrm{d}$
$=8+28 \times 2$
$ =8+56=64$
$a_{29}=64$

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9. If the 3rd and the 9th terms of an AP are 4 and $-8$ respectively, which term of this AP is zero?

Answer

Given, $ \mathrm{a}_{3}=4 $ and $ \mathrm{a}_{9}=-8 $
$a_{3}=a+2 d=4$
$a_{9}=a+8 d=-8$
Subtracting 3rd term from 9th term, we get;
$a+8 d-a-2 d=-8-4=-12$
$6 d=-12$
$d=-2$
Substituting the value of $ d $ in 3rd term, we get;
$a+2(-2)=4$
$a-4=4$
$a=8$
Now; $ 0=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
$0=8+(\mathrm{n}-1)(-2)$
$(\mathrm{n}-1)(-2)=-8$
$\mathrm{n}-1=4$
$\mathrm{n}=5$
Thus, 5th term of this AP is zero.

10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Answer

To find: Common Difference, d nth term of an AP is given by, $ a_{n}=a+(n-1) d $ where, $ a $ is the first term, $ n $ is the number of terms in an AP and $ d $ is the common difference. Tenth and seventeenth terms of this AP can be given as follows:
$ \mathrm{a}_{10}=\mathrm{a}+9 \mathrm{d} $
$ a_{17}=a+16 d $
It is give that 17th term is 7 more than the tenth term, So
Subtracting 10th term from 17th term, we get;
$a+16 d-a-9 d=7$
$7 d=7$
$d=1$
Hence common difference of the A.P is 1.

11. Which term of the $ \mathrm{AP}: 3,15,27,39, \ldots $ will be 132 more than its 54th term?

Answer

To find : n such that $ \mathrm{a}_{\mathrm{n}}=\mathrm{a}_{54}+132 $
Given: $ \mathrm{a}=3, \mathrm{d}=15-3=12 $
nth term of an AP is given by:
$ a_{n}=a+(n-1) d $
where, $ \mathrm{a}= $ first term of $ \mathrm{APn}= $ no. of terms of $ \mathrm{APd}= $ common difference of AP
54th term can be given as follows:
$\mathrm{a}_{54}=\mathrm{a}+(54-1) \mathrm{d}$
$=3+53 \times 12$
$=3+636$
$=639$
In question we have to find the term which is 132 more than 639 i.e. 771
Now $ \mathrm{a}_{\mathrm{n}}=771, \mathrm{n}= $ ?
Again applying the formula of nth term
$771=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}$
$771=3+(\mathrm{n}-1) 12$
$(\mathrm{n}-1) 12=771-3=768$
$\mathrm{n}-1=\frac{768 }{ 12}$
$\mathrm{n}-1=64$
$\mathrm{n}=65$
Thus, the required term is 65th term.

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12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Answer

Let the common difference of two $AP^{\prime} s$ be d, their first terms as a and $ \mathrm{a}^{\prime} $
nth term of both the AP's will be given by
$a_{n}=a+(n-1) d$
$a_{n}^{\prime}=a^{\prime}+(n-1) d$
Now 100th term of 1st AP will be given by: $ a_{100} $
$ =\mathrm{a}+(100-1) \mathrm{d}=\mathrm{a}+99 \mathrm{d} $
100th term of second AP will be given by: $ \mathrm{a}^{\prime}{ }_{100} $
$ =a^{\prime}+(100-1) d=a^{\prime}+99 d $
Given, $ a_{100}-a^{\prime}{ }_{100}=(a+99 d)-\left(a^{\prime}+99 d\right) $
$ \Rightarrow \mathrm{a}_{100}-\mathrm{a}^{\prime}{ }_{100}=\left(\mathrm{a}-\mathrm{a}^{\prime}\right) $
So, difference does not depend on number of terms. Thus, $ \mathrm{a}_{1000}- \mathrm{a}^{\prime}{ }_{1000}=100=\mathrm{a}_{100}-\mathrm{a}^{\prime}{ }_{100} $
So the difference between their 1000th terms is 100.

13. How many three-digit numbers are divisible by 7 ?

Answer

Since, 100 is the smallest three digit number and it gives a reminder of 2 when divided by 7, therefore, 105 is the smallest three digit number which is divisible by 7
Since, 999 is greatest three digit number, and it gives a remainder of 5, thus $ 999-5=994 $ will be the greatest three digit number which is divisible by 7
Therefore, here we have,
First term $ (a)=105 $,
The last term $ \left(a_{n}\right)=994 $
The common difference $ =7 $
We know that, $ \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
Or, $ 994=105+(\mathrm{n}-1) 7 $
Or, $ (\mathrm{n}-1) 7=994-105=889 $
Or, $ \mathrm{n}-1=127 $
Or, $ \mathrm{n}=128 $
Thus, there are 128 three digit numbers which are divisible by 7.

14. How many multiples of 4 lie between 10 and 250 ?

Answer

12 is the first number after 10 which is divisible by 4
Since, 250 gives a remainder of 2 when divided by 4, thus $ 250-2= $ is the greatest number less than 250 which is div 248isible by 4.
Here, we have first term (a) = 12, last term (n) = 248 and common difference $ (d)=4 $
Thus, number of terms $ (\mathrm{n})= $ ?
We know that, an $ =\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
Or, $ 248=12+(\mathrm{n}-1) 4 $
Or, $ (\mathrm{n}-1) 4=248-12=236 $
Or, $ \mathrm{n}-1=59 $
Or, $ \mathrm{n}=60 $
Thus, there are 60 numbers between 10 and 250 that are divisible by 4.

15. For what value of $ n $, are the nth terms of two APs: $ 63,65,67, \ldots $ . and $ 3,10,17, \ldots $ equal?

Answer

In first AP: $ a=63, \mathrm{~d}=2 $
In second AP: $ \mathrm{a}=3, \mathrm{~d}=7 $
As per question:
nth term of first $ \mathrm{AP}= $ nth term of second AP
$ 63+(n-1) 2=3+(n-1) 7 $
$\Rightarrow 63-3+(\mathrm{n}-1) 2=(\mathrm{n}-1) 7$
$\Rightarrow 60+2 \mathrm{n}-2=7 \mathrm{n}-7$
$\Rightarrow 2 \mathrm{n}+58=7 \mathrm{n}-7$
$\Rightarrow 2 \mathrm{n}+58+7=7 \mathrm{n}$
$\Rightarrow 2 \mathrm{n}+65=7 \mathrm{n}$
$\Rightarrow 7 \mathrm{n}-2 \mathrm{n}=65$
$\Rightarrow 5 \mathrm{n}=65$
$\Rightarrow \mathrm{n}=\frac{65 }{ 5}=13$
Thus, for the 13 value of $ n $, nth term of given two APs will be equal.

16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12

Answer

To Find: A.P
Given: $ \mathrm{a}_{3}=16, \mathrm{a}_{7}-\mathrm{a}_{5}=12 $
For this question, We need to find third, fifth and seventh term of an AP by formula of nth term. We know that, nth term of an AP is given by $ : a_{n}=a+(n-1) d $
So, Given $ \mathrm{a}_{3}=16 $ and $ \mathrm{a}_{7}-\mathrm{a}_{5}=12 $
where $ \mathrm{a}_{3}= $ third term of the AP, and so on
$a_{3}=a+2 d=16\quad \ldots\ldots\text{(i)}$
$a_{5}=a+4 d$
$a_{7}=a+6 d$
As per question;
7th term exceeds the fifth term by 12, So the difference of seventh and fifth term will be 12
$a+6 d-a-4 d=12$
$\Rightarrow 2 d=12$
$\Rightarrow d=6$
Substituting the value of $ d $ in eq (i), we get;
$ a+2 \times 6=16 $
Or, $ \mathrm{a}+12=16 $
Or, $ \mathrm{a}=16-12=4 $
Thus, the AP can be given as follows:
$ a, a+d, a+2 d, a+3 d, a+4 d\ldots $ and thus,
$ 4,10,16,22,28, \ldots $

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17. Find the 20th term from the last term of the AP : $ 3,8,13, \ldots , 253$

Answer

To find: 20th term from the last
In the given AP First term, $ a=3 $ common difference, $ d=5 $
we know, nth term of an $ A P $ is $ a_{n}=a+(n-1) d $
Now, Let the no of terms in given AP is n
last term $ =253 $
$\Rightarrow 253=a+(n-1) d$
$\Rightarrow 253=3+(n-1) \times 5$
$\Rightarrow 253=3+5 n-5=-2$
$\Rightarrow 5 n=253+2=255$
$\Rightarrow n=\frac{255 }{ 5}=51$
Total number of terms in AP is 51
If there are $ n $ terms in an $ A P $ then some mth term from end will be equal to $ \mathrm{n}-\mathrm{m}+1 $ term from beginning
Therefore, 20th term from the last term will be 32th term from starting
(as, $ 51-20+1=32 $)
$\mathrm{a}_{32}=\mathrm{a}+31 \mathrm{~d}$
$=3+31 \times 5$
$=3+155=158$
Thus, required term is 158.

18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Answer

Given, $ \mathrm{a}_{8}+\mathrm{a}_{4}=24 $ and $ \mathrm{a}_{10}+\mathrm{a}_{6}=44 $
we know, nth term of an AP is a $ +(n-1) d $
where a is first term and d is common difference. Therefore,
$\mathrm{a}_{8}=\mathrm{a}+7 \mathrm{d}$
$\mathrm{a}_{4}=\mathrm{a}+3 \mathrm{d}$
As per question:
$ a+7 d+a+3 d=24 $
Or, $ 2 \mathrm{a}+10 \mathrm{~d}=24 $
Or, $ a+5 d=12 \quad \ldots\ldots\text{(1)}$
Also,
$a_{10}=a+9 d$
$a_{6}=a+5 d$
Acc. to question :
$a+9 d+a+5 d=44$
$\Rightarrow 2 a+14 d=44$
$\Rightarrow a+7 d=22 \quad \ldots\ldots\text{(2)}$
Subtracting equation (1) from equation (2);
$a+7 d-a-5 d=22-12$
$\Rightarrow 2 d=10 \Rightarrow d=5$
Putting the value of $ d $ in equation (1), we get $ a+5(5)=12 \Rightarrow a+25= $ $ 12 \Rightarrow \mathrm{a}=-13 $
As, first three terms of any AP are $ \mathrm{a}, \mathrm{a}+\mathrm{d}, \mathrm{a}+2 \mathrm{d} $.
First three terms of given AP are $ -13,-13+5=-8,-13+2(5)=-3 $.

19. SubbaRao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000 ?

Answer

Here, $ a=5000, d=200 $ and an $ =7000 $
We know, an $ =\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
$7000=5000+(n-1) 200$
$(n-1) 200=7000-5000$
$(n-1) 200=2000$
$n-1=10$
$n=11$
i.e. after 11 years, starting from 1995, his salary will reach to 7000, so we have to add 10 in 1995, because these numbers are in years.
Thus, $ 1995+10=2005 $
Hence, his salary reached at Rs. 7000 in 2005.

20. Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find $ n $.

Answer

Ramkali's saving in First week = Rs. 5
Ramkali's saving in second week $=$ Ramkali's saving of first week $+$ Ramkali's saving of second week
$ =5+1.75=\text { Rs. } 6.75 $
Rampkali's saving in third week $ =6.75+1.75= $ Rs. 8.50
And thus an AP will be formed as $ 5,6.75 .8 .50, \ldots\ldots $ with common difference of 1.75
Now, Saving in nth week $ = $ Rs. 20.75
Therefore, $ a=5, d=1.75 $ and $ \mathrm{a}_{\mathrm{n}}=20.75 $
We know, $ \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} $
$20.75=5+(n-1) 1.75$
$(\mathrm{n}-1) 1.75=20.75-5$
$(\mathrm{n}-1) 1.75=15.75$
$\mathrm{n}-1=\frac{15.75 }{ 1.75}=9$
$\mathrm{n}=10$
So in 10 weeks her savings will be Rs. 20.75.

CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)

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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)

Download the Math Ninja App Now