NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2 - Arithmetic Progression

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 $20.75$.

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2

The best part about the Maths 5.2 Class 10 PDF solution is that it can be downloaded for free. All that the students need is an internet connection to download the NCERT solutions for Class 10 Maths EX 5.2 from the main website of Vedantu. Once downloaded, they can refer to the solution on the go. The 10 Class Math exercise 5.2 solution can be referred to just before the examination. Students can also print a hard copy of the Exercise 5.2 Maths Class 10 solutions if they want to keep it handy always.

Important Topics under NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.2

NCERT Class 10 Maths Chapter 5 is on arithmetic progressions. This chapter, Arithmetic Progression, has 5 major parts that need to be covered to understand the concepts of arithmetic progression properly. The following table lists the 5 important topics that are covered under the chapter on Arithmetic Progression. We advise students to go through these individual topics carefully to get a precise understanding of the chapter.

Importance of Arithmetic Progressions

An arithmetic progression is basically a sequence of numbers where every number differs from its previous one by a specific quantity. The importance of arithmetic progressions lies in handing out the ability to work closely with patterns and systems, be it in school-level maths or in real life. Students are encouraged to learn all about arithmetic progressions in this chapter to be able to easily solve objective-type problems and word problems based on the same in their exams.

NCERT Solutions for Class 10 Maths Exercise 5.2 – Arithmetic Progression

5.2 Arithmetic Progression

Be it in nature or any company figures you will notice that data follows a particular pattern. This could be anything be it the pattern of the petals on a sunflower or the order in which the honeybee comb is designed and formed. When you notice a pattern you will notice that there is a formula between the succeeding and preceding term. It could be that the succeeding and the preceding terms are the sums or the difference of some number that follows a particular ratio. There are many such progressions when you go about studying mathematics. A very important one among them is the arithmetic progression.

Arithmetic progression or AP is a series where a pattern is formed when the succeeding term is formed by adding a particular number to the previous value. For example, if the first term is ‘a’ then and the next term is ‘a+d’, the number after that is ‘a+2d’ and then ‘a+3d’ and so on. When you notice a pattern of numbers that has d as the common difference between its two consecutive numbers, then this is an arithmetic progression.

nth Term of an AP

If you have been given a series and you figure out that it is in an arithmetic progression, then how do you go about finding out what could be the nth term of this series. This section helps students to derive the formula for the nth term of an AP series that in turn lets you find the nth term of any AP series.

The nth term of the AP series is calculated as: a(n) = a+ (n-1)d

Here ‘a’ is the first term of the series, ‘n’ is the term that you wish to find out, and ‘d’ is the common difference between two successive terms in the AP series. Students need to remember this formula and then need to find out what the n and the d values are. Apply it in the formula and get the nth term of the AP series.

Sum of the First N Terms of an AP

The section explains how on getting an AP series; you can go ahead and find the sum of n terms of an AP series. This is important to know because it makes no sense to use a calculator and add each term of the AP series to find out its sum. The section helps students to derive the formula of the n terms of an AP, and this formula can be used to find out the summation. All that the student needs to do is to remember the formula and then to apply the value correctly to the formula to get the right answers.

The sum of n terms an AP series is calculated as:

S = (n/2) (2a+ (n-1) d)

Also if you have an AP series where the first number is 1, then the sum of n terms of the particular AP series with the first number as one is calculated as:

S = (n/2) (a+1)

Summary

The section summarizes all the important formulas and explains how an AP pattern is formed and how to calculate the nth term and the sum of n terms of an AP series.

Key Features of NCERT Solutions for Class 10 Maths Exercise 5.2

The key features of Class 10 Maths Ex 5.2 solutions are:

• The concept has been explained concisely and solves all the doubts that the student may have when he goes through the chapter.

• The Class 10 Maths Exercise 5.2 Solutions problems are solved through various approaches that let students tackle all kinds of questions on this topic.

• The language used in Exercise 5.2 Class 10 Solutions is easy, which means that anyone can go through these solutions and understand the topic.

NCERT Solutions for Class 10 Maths Chapter 5 Exercises