CBSE Class 10 Maths Chapter 5 Arithmetic Progressions – NCERT Solutions 2025–26

Exercise 5.2

1.  Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and \[{{a}_{n}}\] the \[{{n}^{th}}\] term of the A.P.

Ans: 

i. Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, \[d=3\] …..(2)

Given, the number of Terms, \[n=8\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

${{a}_{n}}=7+\left( 8-1 \right)3$

$\Rightarrow {{a}_{n}}=7+21$

$\therefore {{a}_{n}}=28$

ii. Given, the first Term, $a=-18$  ….. (1)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=0\] …..(2)

Given, the number of Terms, \[n=10\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$0=-18+\left( 10-1 \right)d$

$\Rightarrow 18=9d$

$\therefore d=2$

iii. Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-5\] ….. (1)

Given, the common Difference, \[d=-3\] …..(2)

Given, the number of Terms, \[n=18\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$-5=a+\left( 18-1 \right)\left( -3 \right)$

$\Rightarrow -5=a-51$

$\therefore a = 46$

iv. Given, the first Term, $a=-18.9$  ….. (1)

Given, the common Difference, \[d=2.5\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=3.6\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$3.6=-18.9+\left( n-1 \right)\left( 2.5 \right)$

$\Rightarrow 22.5=\left( n-1 \right)\left( 2.5 \right)$

$\Rightarrow 9=\left( n-1 \right)$

$\therefore n=10$

v. Given, the first Term, $a=3.5$  ….. (1)

Given, the common Difference, \[d=0\] …..(2)

Given, the number of Terms, \[n=105\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

${{a}_{n}}=3.5+\left( 105-1 \right)\left( 0 \right)$

$\therefore {{a}_{n}}=3.5$ 

2. Choose the correct choice in the following and justify

i. \[{{30}^{th}}\] term of the A.P \[10,7,4,...,\] is

1. \[97\]  

2. \[77\]  

3. \[- 77\] 

4. \[87\] 

Ans: Option C. $-77$ 

Given, the first Term, $a=10$  ….. (1)

Given, the common Difference, \[d=7-10=-3\] …..(2)

Given, the number of Terms, \[n=30\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{a}_{30}}=10+\left( 30-1 \right)\left( -3 \right)$

$\Rightarrow {{a}_{30}}=10-87$

$\therefore {{a}_{30}}=-77$

ii. \[{{11}^{th}}\] term of the A.P \[-3,-\dfrac{1}{2},2,...,\] is

1. \[28\]

2. \[22\]  

3. \[38\] 

4. \[48\dfrac{1}{2}\] 

Ans: Option II, $22$ 

Given, the first Term, $a=-3$  ….. (1)

Given, the common Difference, \[d=-\dfrac{1}{2}-\left( -3 \right)=\dfrac{5}{2}\] …..(2)

Given, the number of Terms, \[n=11\]  …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, ${{a}_{n}}=-3+\dfrac{5}{2}\left( 11-1 \right)$

$\Rightarrow {{a}_{n}}=-3+25$

$\therefore {{a}_{n}}=22$ 

3. In the following APs find the missing term in the blanks

i. \[2,\_\_,26\] 

Ans: Given, first term $a=2$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     

Substituting the values from (1) we get, ${{a}_{n}}=2+\left( n-1 \right)d$  …..(2)

Given, third term ${{a}_{3}}=26$. From (2) we get,

$26=2+\left( 3-1 \right)d$

$\Rightarrow 26=2+2d$ 

$\therefore d=12$  ….(3)

From (1), (2) and (3) we get for $n=2$ 

${{a}_{2}}=2+\left( 2-1 \right)\left( 12 \right)$

$\therefore {{a}_{2}}=14$

$\therefore $ The sequence is \[2,14,26\].

ii. \[\_\_,13,\_\_,3\] 

Ans: Given, second term ${{a}_{2}}=13$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ …..(2)

Substituting the values from (1) for $n=2$ we get, $13=a+d$  …..(3)

Given, fourth term ${{a}_{4}}=3$. From (2) we get, 

$3=a+3d$  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

$3-13=\left( a+3d \right)-\left( a+d \right)$

$\Rightarrow -10=2d$ 

$\therefore d=-5$  ….(5)

From (3) and (5) we get 

$13=a-5$

$\Rightarrow a=18$ ……(6)

Substituting the values from (5) and (6) in (2) we get,

${{a}_{n}}=18-5\left( n-1 \right)$   …..(7)

First term, $a=18$ and third term ${{a}_{3}}=8$

$\therefore $ The sequence is \[18,13,8,3\].

iii. $5,\_\_,\_\_,9\dfrac{1}{2}$ 

Ans: Given, first term $a=5$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ …..(2)

Substituting the values from (1) in (2) we get, ${{a}_{n}}=5+\left( n-1 \right)d$  …..(3)

Given, fourth term ${{a}_{4}}=9\dfrac{1}{2}$. From (3) we get,

$9\dfrac{1}{2}=5+\left( 4-1 \right)d$ 

$\Rightarrow 9\dfrac{1}{2}=5+3d$ 

$\therefore d=\dfrac{3}{2}$  ….(4)

From (3) and (4) we get 

${{a}_{n}}=5+\dfrac{3}{2}\left( n-1 \right)$ ……(5)

Second term, ${{a}_{2}}=\dfrac{13}{2}$ and third term ${{a}_{3}}=8$

$\therefore $ The sequence is $5,\dfrac{13}{2},8,9\dfrac{1}{2}$.

iv. $-4,\_\_,\_\_,\_\_,\_\_,6$ 

Ans: Given, first term $a=-4$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ …..(2)

Substituting the values from (1) in (2) we get, ${{a}_{n}}=-4+\left( n-1 \right)d$  …..(3)

Given, sixth term ${{a}_{6}}=6$. From (3) we get, 

$6=-4+\left( 6-1 \right)d$ 

$\Rightarrow 6=-4+5d$ 

$\therefore d=2$  ….(4)

From (3) and (4) we get 

${{a}_{n}}=-4+2\left( n-1 \right)$ ……(5)

Second term ${{a}_{2}}=-2$, third term ${{a}_{3}}=0$, fourth term ${{a}_{4}}=2$ and fifth term ${{a}_{5}}=4$.

$\therefore $ The sequence is $-4,-2,0,2,4,6$

v. $\_\_,38,\_\_,\_\_,\_\_,-22$ 

Ans: Given, second term ${{a}_{2}}=38$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$ …..(2)

Substituting the values from (1) for $n=2$ we get, $38=a+d$  …..(3)

Given, sixth term ${{a}_{6}}=-22$. From (2) we get,

$-22=a+5d$  …..(4)

Solving (3) and (4) by subtracting (3) from (4) we get,

$-22-38=\left( a+5d \right)-\left( a+d \right)$

$\Rightarrow -60=4d$ 

$\therefore d=-15$  ….(5)

From (3) and (5) we get 

$38=a-15$

$\Rightarrow a=53$ ……(6)

Substituting the values from (5) and (6) in (2) we get,
${{a}_{n}}=53-15\left( n-1 \right)$   …..(7)

The first term, $a=53$, second term ${{a}_{3}}=23$, third term ${{a}_{3}}=8$ and fourth term ${{a}_{4}}=-7$

$\therefore $ The sequence is \[53,38,23,8,-7,-22\].

4. Which term of the A.P. \[3,8,13,18,...\] is \[78\]?

Ans: Given, the first Term, $a=3$  ….. (1)

Given, the common Difference, \[d=8-3=5\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=78\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$78=3+5\left( n-1 \right)$

$\Rightarrow 75=5\left( n-1 \right)$

$\Rightarrow 15=\left( n-1 \right)$

$\therefore n=16$ 

Therefore, \[{{16}^{th}}\] term of this A.P. is \[78\].

5. Find the number of terms in each of the following A.P.

i. \[\text{7,13,19,}...\text{,205}\]

Ans: Given, the first Term, $a=7$  ….. (1)

Given, the common Difference, \[d=13-7=6\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=205\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$205=7+6\left( n-1 \right)$

$\Rightarrow 198=6\left( n-1 \right)$

$\Rightarrow 33=\left( n-1 \right)$

$\therefore n=34$ 

Therefore, given A.P. series has \[34\] terms.

ii. \[18,15\dfrac{1}{2},13,....,-47\] 

Ans: Given, the first Term, $a=18$  ….. (1)

Given, the common Difference, \[d=15\dfrac{1}{2}-18=-\dfrac{5}{2}\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-47\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$-47=18-\dfrac{5}{2}\left( n-1 \right)$

$\Rightarrow -65=-\dfrac{5}{2}\left( n-1 \right)$

$\Rightarrow 26=\left( n-1 \right)$

$\therefore n=27$ 

Therefore, given A.P. series has \[27\] terms.

6. Check whether \[-150\] is a term of the A.P. \[11,8,5,2,...\] 

Ans: Given, the first Term, $a=11$  ….. (1)

Given, the common Difference, \[d=8-11=-3\] …..(2)

Given, the ${{n}^{th}}$ term, \[{{a}_{n}}=-150\] …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get,

$-150=11-3\left( n-1 \right)$

$\Rightarrow -161=-3\left( n-1 \right)$

$\Rightarrow \dfrac{161}{3}=\left( n-1 \right)$

$\therefore n=\dfrac{164}{3}$ 

Since $n$ is nor a natural number. Therefore, $-150$ is not a term of the given A.P. series.

7. Find the \[{{31}^{st}}\] term of an A.P. whose \[{{11}^{th}}\] term is \[38\] and the \[{{16}^{th}}\] term is \[73\].

Ans: Given, the \[{{11}^{th}}\] Term, ${{a}_{11}}=38$  ….. (1)

Given, the \[{{16}^{th}}\] Term, ${{a}_{16}}=73$  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(3)

Substituting the values from (1) in (3) we get,

$38=a+\left( 11-1 \right)d$

$\Rightarrow 38=a+10d$      …..(4)

Substituting the values from (2) in (3) we get,

$73=a+\left( 16-1 \right)d$

$\Rightarrow 73=a+15d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow 73-38=\left( a+15d \right)-\left( a+10d \right)$

$\Rightarrow 5=35d$

$\therefore d=7$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 38=a+70$

$\therefore a=-32$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=-32+7\left( n-1 \right)$  …..(8)

To find the ${{31}^{st}}$ term substitute $n=31$ in (8) we get, 

${{a}_{31}}=-32+7\left( 31-1 \right)$

$\Rightarrow {{a}_{31}}=-32+210$

$\therefore {{a}_{31}}=178$

Therefore, the \[{{31}^{st}}\] term of an A.P. whose \[{{11}^{th}}\] term is \[38\] and the \[{{16}^{th}}\] term is \[73\] is $178$. 

8. An A.P. consists of \[50\] terms of which \[{{3}^{rd}}\] term is \[12\] and the last term is \[106\]. Find the \[{{29}^{th}}\] term.

Ans: Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=12$  ….. (1)

Given, the \[{{50}^{th}}\] Term, ${{a}_{50}}=106$  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(3)

Substituting the values from (1) in (3) we get,

$12=a+\left( 3-1 \right)d$

$\Rightarrow 12=a+2d$      …..(4)

Substituting the values from (2) in (3) we get,

$106=a+\left( 50-1 \right)d$

$\Rightarrow 106=a+49d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow 106-12=\left( a+49d \right)-\left( a+2d \right)$

$\Rightarrow 94=47d$

$\therefore d=2$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 12=a+4$

$\therefore a=8$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=8+2\left( n-1 \right)$  …..(8)

To find the ${{29}^{th}}$ term substitute $n=29$ in (8) we get, 

${{a}_{29}}=8+2\left( 29-1 \right)$

$\Rightarrow {{a}_{29}}=8+56$

$\therefore {{a}_{29}}=64$

Therefore, the ${{29}^{th}}$ term of the A.P. is $64$. 

9. If the \[{{3}^{rd}}\] and the \[{{9}^{th}}\] terms of an A.P. are $4$ and \[8\] respectively. Which term of this A.P. is zero.

Ans: Given, the \[{{3}^{rd}}\] Term, ${{a}_{3}}=4$  ….. (1)

Given, the \[{{9}^{th}}\] Term, ${{a}_{9}}=-8$  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     …..(3)

Substituting the values from (1) in (3) we get,

$4=a+\left( 3-1 \right)d$

$\Rightarrow 4=a+2d$      …..(4)

Substituting the values from (2) in (3) we get,

$-8=a+\left( 9-1 \right)d$

$\Rightarrow -8=a+8d$      …..(5)

Solving equations (4) and (5) by subtracting (4) from (5) we get,

$\Rightarrow -8-4=\left( a+8d \right)-\left( a+2d \right)$

$\Rightarrow -12=6d$

$\therefore d=-2$   …..(6)

Substituting value from (6) in (4) we get,

$\Rightarrow 4=a-4$

$\therefore a=8$   …..(7)

Again, substituting the values from (6) and (7) in (3) we get,

${{a}_{n}}=8-2\left( n-1 \right)$  …..(8)

T find the term which is zero, substitute ${{a}_{n}}=0$ in (8)

$0=8-2\left( n-1 \right)$

$\Rightarrow 8=2\left( n-1 \right)$

$\Rightarrow 4=\left( n-1 \right)$

$\therefore n=5$ 

Therefore, given A.P. series has \[{{5}^{th}}\] term as zero.

10. If \[{{17}^{th}}\] term of an A.P. exceeds its \[{{10}^{th}}\] term by \[7\]. Find the common difference.

Ans: Given that the \[{{17}^{th}}\] term of an A.P. exceeds its \[{{10}^{th}}\] term by \[7\] i.e., 

${{a}_{17}}={{a}_{10}}+7$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (2)

For \[{{17}^{th}}\] term substitute $n=17$ in (2) i.e., ${{a}_{17}}=a+16d$  ….. (3)

For \[{{10}^{th}}\] term substitute $n=10$ in (2) i.e., ${{a}_{10}}=a+9d$  ….. (4)

Therefore, from (1), (3) and (4) we get, 

$a+16d=a+9d+7$

$\Rightarrow 7d=7$

$\therefore d=1$ 

Therefore, the common difference is $1$.

11. Which term of the A.P. \[3,15,27,39,...\] will be \[132\] more than its \[{{54}^{th}}\] term?

Ans: Let ${{n}^{th}}$ term of A.P. be \[132\] more than its \[{{54}^{th}}\] term i.e.,

${{a}_{n}}={{a}_{54}}+132$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (2)

For \[{{54}^{th}}\] term substitute $n=54$ in (2) i.e., ${{a}_{54}}=a+53d$  ….. (3)

Therefore, from (1), (2) and (3) we get, 

$a+\left( n-1 \right)d=a+53d+132$

$\Rightarrow \left( n-1 \right)d-53d=132$

$\therefore d=\dfrac{132}{n-54}$  ….. (4)

Now, given A.P. \[3,15,27,39,...\] 

Common difference $d=15-3=12$  ….. (5)

Hence, from (4) and (5) we get $12=\dfrac{132}{n-54}$

$\Rightarrow n-54=11$ 

$\therefore n=65$ 

Therefore, ${{65}^{th}}$ term of the given A.P. will be \[132\] more than its \[{{54}^{th}}\] term.

12. Two APs have the same common difference. The difference between their \[{{100}^{th}}\] term is \[100\], what is the difference between their \[{{1000}^{th}}\] terms?

Ans: Let $2$ A.P.’s be 

$a,a+d,a+2d,a+3d,....$   …..(1)

$b,b+d,b+2d,b+3d,....$    …..(2)

(Since common difference is same)

Given that the difference between their \[{{100}^{th}}\] term is \[100\] i.e., 

${{a}_{100}}-{{b}_{100}}=100$ …..(3)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (4)

Therefore, from (3) and (4) we get, 

$a+\left( 100-1 \right)d-\left( b+\left( 100-1 \right)d \right)=100$

$\Rightarrow a-b=100$  ….. (5)

Similarly, the difference between their \[{{1000}^{th}}\] terms is, 

${{a}_{1000}}-{{b}_{1000}}=\left[ a+\left( 1000-a \right)d \right]-\left[ b+\left( 1000-a \right)d \right]$

$\Rightarrow {{a}_{1000}}-{{b}_{1000}}=a-b$

$\therefore {{a}_{1000}}-{{b}_{1000}}=100$ 

Therefore, the difference between their \[{{1000}^{th}}\] terms is $100$.

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

Ans: First three-digit number that is divisible by $7$ is \[105\] then the next number will be \[105+7=112\]. 

Therefore, the series becomes \[105,112,119,....\] 

This is an A.P. having first term as \[105\] and common difference as \[7\].

Now, the largest $3$ digit number is $999$.

Leu us divide it by \[7\] , to get the remainder.

$999=142\times 7+5$ 

Therefore, \[999-5=994\] is the maximum possible three-digit number that is divisible by $7$.

Also, this will be the last term of the A.P. series.

Hence the final series is as follows: \[105,112,119,...,994\] 

Let 994 be the \[{{n}^{th}}\] term of this A.P.

Then, \[{{a}_{n}}=105+7\left( n-1 \right)\] 

\[\Rightarrow 994=105+7\left( n-1 \right)\]

\[\Rightarrow 889=7\left( n-1 \right)\]

\[\Rightarrow 127=\left( n-1 \right)\]

$\therefore n=128$ 

Therefore, $128$ three-digit numbers are divisible by $7$.

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

Ans: First number that is divisible by $4$ and lie between $10$ and $250$ is $12$. The next number will be \[12+4=16\]. 

Therefore, the series becomes \[12,16,20,....\] 

This is an A.P. having first term as \[12\] and common difference as \[4\].

Now, the largest number in range is $250$.

Leu us divide it by \[4\] to get the remainder.

$250=62\times 4+2$ 

Therefore, \[250-2=248\] is the last term of the A.P. series.

Hence the final series is as follows: \[12,16,20,....,248\] 

Let $248$ be the \[{{n}^{th}}\] term of this A.P.

Then, \[{{a}_{n}}=12+4\left( n-1 \right)\] 

\[\Rightarrow 248=12+4\left( n-1 \right)\]

\[\Rightarrow 236=4\left( n-1 \right)\]

\[\Rightarrow 59=\left( n-1 \right)\]

$\therefore n=60$ 

Therefore, $60$ multiples of $4$ lie between $10$ and $250$.

15. For what value of $n$, are the ${{n}^{th}}$ terms of two APs \[63,65,67,....\] and \[3,10,17,....\] equal

 Ans: Given $2$ A.P.’s are

\[63,65,67,....\]   …..(1)

Its first term is $63$ and common difference is $65-63=2$ 

\[3,10,17,....\]    …..(2)

Its first term is $3$ and common difference is $10-3=7$ 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

Therefore, from (1) and (3) we get the ${{n}^{th}}$ term of the first A.P. is

 ${{a}_{n}}=63+2\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=61+2n$  ….. (4)

And from (2) and (3) we get the ${{n}^{th}}$ term of the second A.P. is

 ${{b}_{n}}=3+7\left( n-1 \right)$

$\Rightarrow {{b}_{n}}=-4+7n$  ….. (5)

If the ${{n}^{th}}$ terms of two APs \[63,65,67,....\] and \[3,10,17,....\] are equal the from (4) and (5),

\[{{a}_{n}}={{b}_{n}}\]

$\Rightarrow 61+2n=-4+7n$

$\Rightarrow 65=5n$

$\therefore n=13$ 

Therefore, the ${{13}^{th}}$ term of both the A.P.’s are equal.

16. Determine the A.P. whose third term is \[16\] and the \[{{7}^{th}}\] term exceeds the \[{{5}^{th}}\] term by $12$.

Ans: Given the ${{7}^{th}}$ term of A.P. is \[12\] more than its \[{{5}^{th}}\] term i.e.,

${{a}_{7}}={{a}_{5}}+12$ …..(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (2)

For \[{{5}^{th}}\] term substitute $n=5$ in (2) i.e., ${{a}_{5}}=a+4d$  ….. (3)

For \[{{7}^{th}}\] term substitute $n=7$ in (2) i.e., ${{a}_{7}}=a+6d$  ….. (4)

Therefore, from (1), (3) and (4) we get, 

$a+6d=a+4d+12$

$\Rightarrow 2d=12$

$\therefore d=6$   ….. (5)

Substituting (5) in (2) we get, ${{a}_{n}}=a+6\left( n-1 \right)$ ……(6)

Given the third term of the A.P. is \[16\]. Hence from (6),

$16=a+6\left( 3-1 \right)$

$\Rightarrow 16=a+12$

$\therefore a=4$   ….. (7)

Hence from (6), ${{a}_{n}}=4+6\left( n-1 \right)$

Therefore, the A.P. will be \[4,10,16,22,....\].

17. Find the \[{{20}^{th}}\] term from the last term of the A.P. \[3,8,13,...,253\] 

Ans: Given A.P. \[3,8,13,...,253\]. To find the \[{{20}^{th}}\] term from the last write the given A.P. in reverse order and then find its \[{{20}^{th}}\] term.

Required A.P. is $253,...,13,8,3$  ….. (1) 

Its first A.P. is $253$ and common difference is $8-13=-5$.  …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

Hence from (2) and (3) we get, ${{a}_{n}}=253-5\left( n-1 \right)$   …..(4)

Substitute $n=20$ in (4) we get, 

${{a}_{20}}=253-5\left( 20-1 \right)$

$\Rightarrow {{a}_{20}}=158$.

Therefore, ${{20}^{th}}$ term from the last is $158$.

18. The sum of \[{{4}^{th}}\] and \[{{8}^{th}}\] terms of an A.P. is \[24\] and the sum of the \[{{6}^{th}}\] and \[{{10}^{th}}\] terms is \[44\]. Find the first three terms of the A.P.  

Ans: Given the sum of \[{{4}^{th}}\] and \[{{8}^{th}}\] terms of an A.P. is \[24\] i.e.,

${{a}_{4}}+{{a}_{8}}=24$ …..(1)

Given the sum of \[{{6}^{th}}\] and \[{{10}^{th}}\] terms is \[44\] i.e.,

${{a}_{6}}+{{a}_{10}}=44$ …..(2)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$     ….. (3)

For \[{{4}^{th}}\] term substitute $n=4$ in (3) i.e., ${{a}_{4}}=a+3d$  ….. (4)

For \[{{6}^{th}}\] term substitute $n=6$ in (3) i.e., ${{a}_{6}}=a+5d$  ….. (5)

For \[{{8}^{th}}\] term substitute $n=8$ in (3) i.e., ${{a}_{8}}=a+7d$  ….. (6)

For \[{{10}^{th}}\] term substitute $n=10$ in (3) i.e., ${{a}_{10}}=a+9d$  ….. (7)

Therefore, from (1), (6) and (4) we get, 

$\left( a+3d \right)+\left( a+7d \right)=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$   ….. (8)

From (2), (5) and (7) we get, 

$\left( a+5d \right)+\left( a+9d \right)=44$

$\Rightarrow 2a+14d=44$

$\Rightarrow a+7d=22$   ….. (9)

Subtracting (8) from (9) we get,  

$\Rightarrow \left( a+7d \right)-\left( a+5d \right)=22-12$

$\Rightarrow 2d=10$

$\therefore d=5$ ….. (10)

Substituting this value from (10) in (9) we get,

$a+35=22$

$\therefore a=-13$ ….. (11)

Therefore from (10) and (11), the first three terms of the A.P. are $-13,-8,-3$.

19. Subba Rao 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 \[7000\]? 

Ans: Given in the first year, annual salary is Rs $5000$.  

In the second year, annual salary is Rs $5000+200=5200$.  

In the third year, annual salary is Rs $5200+200=5400$. 

This series will form an A.P. with first term $5000$ and common difference $200$.  

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$  

Therefore, In the ${{n}^{th}}$ year, annual salary is ${{a}_{n}}=5000+200\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=4800+200n$ …. (1)

To find the year in which his annual income reaches Rs \[7000\], substitute ${{a}_{n}}=7000$ in (1) and find the value of $n$i.e., 

  $7000=4800+200n$

$\Rightarrow 2200+200n$

$\therefore n=11$ 

Therefore, in ${{11}^{th}}$ year i.e., in $2005$ his salary will be Rs $7000$.

20. Ramkali saved Rs \[5\] in the first week of a year and then increased her weekly saving by Rs \[1.75\]. If in the ${{n}^{th}}$ week, her weekly savings become Rs \[20.75\], find $n$. 

Ans: Given in the first week the savings is Rs $5$.  

In the second week the savings is Rs $5+1.75=6.75$.  

In the third week the savings is Rs $6.75+1.75=8.5$. 

This series will form an A.P. with first term $5$ and common difference $1.75$.  

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$  

Therefore, In the ${{n}^{th}}$ week the savings is ${{a}_{n}}=5+1.75\left( n-1 \right)$

$\Rightarrow {{a}_{n}}=3.25+1.75n$ …. (1)

To find the week in which her savings reaches Rs \[20.75\], substitute ${{a}_{n}}=20.75$ in (1) and find the value of $n$i.e., 

  $20.75=3.25+1.75n$

$\Rightarrow 17.5=1.75n$

$\therefore n=10$ 

Therefore, in ${{10}^{th}}$ week her savings will be Rs $20.75$.

Conclusion

Class 10 Ex 5.2 of Maths Chapter 5 - Arithmetic Progressions (AP), is crucial for a solid foundation in math. Understanding the concept of Arithmetic Progression, identifying the common difference, and solving problems using the formula for nth term $a_n = a + (n-1)d$ are key takeaways. Regular practice with Class 10 Maths Exercise 5.2 NCERT solutions provided by platforms like Vedantu can enhance comprehension and problem-solving skills. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in AP-related problems.

Class 10 Maths Chapter 5: Exercises Breakdown

CBSE Class 10 Maths Chapter 5 Other Study Materials

A Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.