NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

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 $a_n=a+(n-1)d$     …..(4)

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

$a_n=7+(8-1)3$

$\Rightarrow a_n=7+21$

$\therefore a_n=28$

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 $a_n=a+(n-1)d$     …..(4)

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

$0=-18+(10-1)d$

$\Rightarrow 18=9d$

$\therefore d=2$

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 $a_n=a+(n-1)d$     …..(4)

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

$-5=a+(18-1)(-3)$

$\Rightarrow -5=a-51$

$\therefore a=46$

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 $a_n=a+(n-1)d$     …..(4)

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

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

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

$\Rightarrow 9=(n-1)$

$\therefore n=10$

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 $a_n=a+(n-1)d$     …..(4)

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

$a_n=3.5+(105-1)\left(0\right)$

$\therefore a_n=3.5$ 

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 $a_n=a+(n-1)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $a_{30}=10+(30-1)(-3)$

$\Rightarrow a_{30}=10-87$

$\therefore a_{30}=-77$

ii. $${11}^{th}$$ term of the A.P $$-3,-{\displaystyle \frac{1}{2}},2,...,$$ is

1. $$28$$

2. $$22$$  

3. $$38$$ 

4. $$48{\displaystyle \frac{1}{2}}$$ 

Ans: Option II, $22$ 

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

Given, the common Difference, $$d=-{\displaystyle \frac{1}{2}}-(-3)={\displaystyle \frac{5}{2}}$$ …..(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 $a_n=a+(n-1)d$     …..(4)

Substituting the values from (1), (2) and (3) in (4) we get, $a_n=-3+{\displaystyle \frac{5}{2}}(11-1)$

$\Rightarrow a_n=-3+25$

$\therefore a_n=22$ 

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

i. $$2,\mathrm{\_}\mathrm{\_},26$$ 

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 $a_n=a+(n-1)d$     

Substituting the values from (1) we get, $a_n=2+(n-1)d$  …..(2)

Given, third term $a_3=26$. From (2) we get,

$26=2+(3-1)d$

$\Rightarrow 26=2+2d$ 

$\therefore d=12$  ….(3)

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

$a_2=2+(2-1)\left(12\right)$

$\therefore a_2=14$

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

ii. $$\mathrm{\_}\mathrm{\_},13,\mathrm{\_}\mathrm{\_},3$$ 

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 $a_n=a+(n-1)d$ …..(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,

$3-13=(a+3d)-(a+d)$

$\Rightarrow -10=2d$ 

$\therefore d=-5$  ….(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,

$a_n=18-5(n-1)$   …..(7)

First term, $a=18$ and third term $a_3=8$

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

iii. $5,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},9{\displaystyle \frac{1}{2}}$ 

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 $a_n=a+(n-1)d$ …..(2)

Substituting the values from (1) in (2) we get, $a_n=5+(n-1)d$  …..(3)

Given, fourth term $a_4=9{\displaystyle \frac{1}{2}}$. From (3) we get,

$9{\displaystyle \frac{1}{2}}=5+(4-1)d$ 

$\Rightarrow 9{\displaystyle \frac{1}{2}}=5+3d$ 

$\therefore d={\displaystyle \frac{3}{2}}$  ….(4)

From (3) and (4) we get 

$a_n=5+{\displaystyle \frac{3}{2}}(n-1)$ ……(5)

Second term, $a_2={\displaystyle \frac{13}{2}}$ and third term $a_3=8$

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

iv. $-4,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},6$ 

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 $a_n=a+(n-1)d$ …..(2)

Substituting the values from (1) in (2) we get, $a_n=-4+(n-1)d$  …..(3)

Given, sixth term $a_6=6$. From (3) we get, 

$6=-4+(6-1)d$ 

$\Rightarrow 6=-4+5d$ 

$\therefore d=2$  ….(4)

From (3) and (4) we get 

$a_n=-4+2(n-1)$ ……(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. $\mathrm{\_}\mathrm{\_},38,\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},\mathrm{\_}\mathrm{\_},-22$ 

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 $a_n=a+(n-1)d$ …..(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,

$-22-38=(a+5d)-(a+d)$

$\Rightarrow -60=4d$ 

$\therefore d=-15$  ….(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,
$a_n=53-15(n-1)$   …..(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 $a_n=a+(n-1)d$     …..(4)

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

$78=3+5(n-1)$

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

$\Rightarrow 15=(n-1)$

$\therefore n=16$ 

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 $a_n=a+(n-1)d$     …..(4)

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

$205=7+6(n-1)$

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

$\Rightarrow 33=(n-1)$

$\therefore n=34$ 

Therefore, given A.P. series has $$34$$ terms.

ii. $$18,15{\displaystyle \frac{1}{2}},13,....,-47$$ 

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

Given, the common Difference, $$d=15{\displaystyle \frac{1}{2}}-18=-{\displaystyle \frac{5}{2}}$$ …..(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 $a_n=a+(n-1)d$     …..(4)

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

$-47=18-{\displaystyle \frac{5}{2}}(n-1)$

$\Rightarrow -65=-{\displaystyle \frac{5}{2}}(n-1)$

$\Rightarrow 26=(n-1)$

$\therefore n=27$ 

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 $a_n=a+(n-1)d$     …..(4)

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

$-150=11-3(n-1)$

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

$\Rightarrow {\displaystyle \frac{161}{3}}=(n-1)$

$\therefore n={\displaystyle \frac{164}{3}}$ 

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 $a_n=a+(n-1)d$     …..(3)

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

$38=a+(11-1)d$

$\Rightarrow 38=a+10d$      …..(4)

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

$73=a+(16-1)d$

$\Rightarrow 73=a+15d$      …..(5)

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

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

$\Rightarrow 5=35d$

$\therefore d=7$   …..(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,

$a_n=-32+7(n-1)$  …..(8)

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

$a_{31}=-32+7(31-1)$

$\Rightarrow a_{31}=-32+210$

$\therefore a_{31}=178$

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 $a_n=a+(n-1)d$     …..(3)

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

$12=a+(3-1)d$

$\Rightarrow 12=a+2d$      …..(4)

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

$106=a+(50-1)d$

$\Rightarrow 106=a+49d$      …..(5)

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

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

$\Rightarrow 94=47d$

$\therefore d=2$   …..(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,

$a_n=8+2(n-1)$  …..(8)

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

$a_{29}=8+2(29-1)$

$\Rightarrow a_{29}=8+56$

$\therefore a_{29}=64$

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 $a_n=a+(n-1)d$     …..(3)

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

$4=a+(3-1)d$

$\Rightarrow 4=a+2d$      …..(4)

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

$-8=a+(9-1)d$

$\Rightarrow -8=a+8d$      …..(5)

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

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

$\Rightarrow -12=6d$

$\therefore d=-2$   …..(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,

$a_n=8-2(n-1)$  …..(8)

T find the term which is zero, substitute $a_n=0$ in (8)

$0=8-2(n-1)$

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

$\Rightarrow 4=(n-1)$

$\therefore n=5$ 

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

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

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

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

Now, given A.P. $$3,15,27,39,...$$ 

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

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

$\Rightarrow n-54=11$ 

$\therefore n=65$ 

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 $a_n=a+(n-1)d$     ….. (4)

Therefore, from (3) and (4) we get, 

$a+(100-1)d-(b+(100-1)d)=100$

$\Rightarrow a-b=100$  ….. (5)

Similarly, the difference between their $${1000}^{th}$$ terms is, 

$a_{1000}-b_{1000}=[a+(1000-a)d]-[b+(1000-a)d]$

$\Rightarrow a_{1000}-b_{1000}=a-b$

$\therefore a_{1000}-b_{1000}=100$ 

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, $$a_n=105+7(n-1)$$ 

$$\Rightarrow 994=105+7(n-1)$$

$$\Rightarrow 889=7(n-1)$$

$$\Rightarrow 127=(n-1)$$

$\therefore n=128$ 

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, $$a_n=12+4(n-1)$$ 

$$\Rightarrow 248=12+4(n-1)$$

$$\Rightarrow 236=4(n-1)$$

$$\Rightarrow 59=(n-1)$$

$\therefore n=60$ 

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 $a_n=a+(n-1)d$     ….. (3)

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

 $a_n=63+2(n-1)$

$\Rightarrow a_n=61+2n$  ….. (4)

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

 $b_n=3+7(n-1)$

$\Rightarrow b_n=-4+7n$  ….. (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 $a_n=a+(n-1)d$     ….. (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, $a_n=a+6(n-1)$ ……(6)

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

$16=a+6(3-1)$

$\Rightarrow 16=a+12$

$\therefore a=4$   ….. (7)

Hence from (6), $a_n=4+6(n-1)$

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 $a_n=a+(n-1)d$     ….. (3)

Hence from (2) and (3) we get, $a_n=253-5(n-1)$   …..(4)

Substitute $n=20$ in (4) we get, 

$a_{20}=253-5(20-1)$

$\Rightarrow a_{20}=158$.

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 $a_n=a+(n-1)d$     ….. (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, 

$(a+3d)+(a+7d)=24$

$\Rightarrow 2a+10d=24$

$\Rightarrow a+5d=12$   ….. (8)

From (2), (5) and (7) we get, 

$(a+5d)+(a+9d)=44$

$\Rightarrow 2a+14d=44$

$\Rightarrow a+7d=22$   ….. (9)

Subtracting (8) from (9) we get,  

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

$\Rightarrow 2d=10$

$\therefore d=5$ ….. (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 $a_n=a+(n-1)d$  

Therefore, In the $n^{th}$ year, annual salary is $a_n=5000+200(n-1)$

$\Rightarrow a_n=4800+200n$ …. (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 $1.75$.