NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 5 exercise 5.2 .
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Complete 10th Class Maths Solution
- NCERT Class 10th – Chapter 4 – Exercise 4.1 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.2 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.3 Quadratic Equations
EXERCISE -5.2
QUE:- 1.Fill in the blanks in the following table, given that a is the first term, d the common
difference and a(n) the nth term of the AP:
| a | d | n | a(n) |
| (i) 7 | 3 | 8 | …….. |
| (ii) -18 | ……. | 10 | 0 |
| (iii) …… | -3 | 18 | -5 |
| (iv) -18.9 | 2.5 | ………. | 3.6 |
| (v) 3.5 | 0 | 105 | ………. |
SOL:-
To solve the question 1 we will use the formula a(n) =a + (n-1)d
(i) a(n) =a + (n-1)d
a(n) =7+ (8-1)3
= 28
(ii) a(n) =a + (n-1)d
0 =-18+ (10-1)d
18 =9d
d=18/9=2
(iii) a(n) =a + (n-1)d
-5 =a + (18-1)-3
a= -51 +5 = -46
(iv) a(n) =a + (n-1)d
3.6 =-18.9 + (n-1)2.5
2.5n =3.6+18.9 +2.5
n =25/2.5=10
(v) a(n) =a + (n-1)d
a(n) =3.5 + (105-1)0
a(n) =3.5
QUE:-2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
(ii) 11th term of the AP: – 3,-1/2 , 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48(1/2)
SOL:-
(i) 30th term of the AP: 10, 7, 4, . . . ,
(A) 97 (B) 77 (C) –77 (D) – 87
a=10 ,d= -3 n=30
a(n) =a + (n-1)d
a(n) =10 + (30-1)(-3)
a(n) =10 -87=-77
ans =(C) -77
(ii) 11th term of the AP: – 3,-1/2 , 2, . . .
(A) 28 (B) 22 (C) –38 (D) – 48(1/2)
a=-3 ,d= 5/2 n=11
a(n) =a + (n-1)d
a(n) =-3+ (11-1)(5/2)
a(n) =-3+25=22
ans =(B) 22
QUE-3. In the following APs, find the missing terms in the boxes :
(i) 2,(…) , 26
(ii) (…), 13,(…) , 3
(iii) 5,(…) ,(…) ,91/2
(iv) – 4,(…) ,(…) ,(…) ,(…) , 6
(v) (…), 38,(…),(…), (…), – 22
SOL:-
(i) 2,(…) , 26
a=2,d= ? n=3 a(n) =26
a(n) =a + (n-1)d
26=2+ (3-1)(d)
d =(26-2)/2=12
a2 = a1 +d =2 + 12 =14
ans =2,(14) , 26
(ii) (…), 13,(…) , 3
a2 =13
a(n) =a + (n-1)d
13 =a + (2-1)d =a +d
a= 13-d ……..(1)
3 =a + (4-1)d =a +3d ………..( 2)
putting the value of a in equ.(2)
a +3d =3
13-d +3d =3
2d = -10
d =-10/2 =-5
putting the value of d in equ. (1)
a = 13 -(-5) = 18
a3 =a +(3-1) (-5)
= 18 -10 =8
ans =(18), 13,(8) , 3
(iii) 5,(…) ,(…) ,9½
a4 =5 +(4-1) (d)
19/2 = 5+3d
19 =10 +6d
d = 9/6 = 3/2
a2 =5 +(2-1)(3/2) =5 +3/2 = 13/2
a3 = 5 + (3-1)(3/2) = 5+ 6/2 =16/2 =8
ans =5,(13/2) ,(8) ,9½
(iv) – 4,(…) ,(…) ,(…) ,(…) , 6
a6 = -4 +(6-1)d
6 = -4 +5d
d =10/5 =2
a2 = -4 +(2-1)2 = -2
a3 = -4 +(3-1)2 =0
a4 = -4 +(4-1)2 =2
a5 = -4 +(5-1)2 =4
ans = – 4,(-2) ,(0) ,(2) ,(4) , 6
(v) (…), 38,(…),(…), (…), – 22
a2 =a +(2-1)d
38 = a + d
d = 38-a……….(1)
a6 = a + (6 -1) d
-22 = a +5d ………..(2)
putting the value of d in equ. (2)
-22 = a +5d
-22 = a +5(38-a) = a +190-5a
-4a = -212
a =-212/-4 =53
a = 53
putting the value of a in equ.(1)
d = 38-a……….(1)
d = 38-53 =-15
a3 = 53 +(3-1)-15=23
a4 = 53 +(4-1)-15=8
a5 = 53 +(5-1)-15=-7
ans = (53), 38,(23),(8), (-7), – 22
QUE:-4. Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
SOL:-
a(n) =a + (n-1)d ; a =3 , d= 5
78=3 + (n-1)5
78 = 3 +5n -5
5n =78 +2
n= 80/5 =16
so 78 is 16th term of AP .
QUE:-5. Find the number of terms in each of the following APs :
(i) 7, 13, 19, . . . , 205 (ii) 18,15½, 13, . . . , – 47
SOL:-
(i) 7, 13, 19, . . . , 205
a(n) =a + (n-1)d ; a =7, d= 6
205 = 7 +(n -1)6
205 = 7 +6n -6
6n =204
n =204/6 =34
number of term in AP is 34
(ii) 18,15½, 13, . . . , – 47
a(n) =a + (n-1)d ; a =18, d= -5/2
-47 = 18 +(n -1)-5/2
-47 = 18 -5/2n -5/2
-47 = 31/2-5/2n
5/2n =31/2 +47
n =( 31+94)×2/2 =125
number of term in AP is 125
QUE:-6. Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
SOL:-
let the -150 is n term of AP
a(n) =a + (n-1)d
-150=11+ (n-1)(-3)
-150 = 11-3n +3
3n = 14+150 =164
n =164/3 =54.666
n is never be in decimal therefore -150 is not a term of AP
QUE:-7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
SOL:-
a11 = a + (11-1)d = a +10d
38 = a +10d
a = 38-10d…………..(1)
a16 = a +(16-1)d
73 = a +15d ……….(2)
putting the value of a in equ. (2)
73 = 38-10d+15d
73 = 38 +5d
5d =73-38 =35
d = 35/5 =7
putting the value of d in equ.(1)
a = 38-10×7
a = -32
a31 = -32 + (31-1)7 = -32 +210=188
so 31st term of AP is 188 .
QUE:-8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th
term.
SOL:-
a50 =106
a3 = 12
a50 =a +(50-1)d
106 = a +49d
a = 106 -49d …………..(1)
a3 = a +(3-1) d
12 = a +2d ……………..(2)
putting the value of a in equ. (2)
12 = 106-49d +2d
-47d =12-106
d =-94/-47 =2
putting the value of d in equ(1)
a = 106 -49×2
a = 106-98 =8
a29 = 8 +(29-1)2 =8+56=64
so 29th term of AP is 64 .
QUE:-9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is
zero?
SOL:-
a3 =4 ; a9 = -8
a3 =a +(3-1)d
4 = a +2d
a = 4-2d …….(1)
a9 = a + (9-1)d
-8 = a +8d …….(2)
putting the value of a in equ.(2)
-8 =4-2d+8d = 4+6d
6d = -12
d =-2
putting the value of d in equ(1)
a = 4-2×-2 =8
a(n) = a +(n-1)d
0 = 8 +(n-1)-2
0 =8 -2n +2
-2n =-10
n =-10/-2=5
so the 0 is 5th term of AP
QUE:-10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
SOL:-
a17 =a +(n-1)d ; a+(17-1)d
a17 =a +16d
a10 =a+(10-1)d =a+9d
according to the question
a17 =a10 +7
a +16d =a+9d +7
16d -9d =7
7d =7
d =7/7 =1
so the common diffrence 1 for AP .
QUE:-11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
SOL:-
a54 =a +(54-1) d ; a =3 ,d =12
a54 = 3 +53× 12
= 3 +636 =639
according to the question
a54 + 132 = a+(n-1)d
639+132 = 3 + (n -1)12
771 =3 +12n-12
771 = 12n -9
771 +9 =12n
12n =780
n=780/12=65
so the 65th term of AP is 132 more than 54th term .
QUE:-12. Two APs have the same common difference. The difference between their 100th terms is
100, what is the difference between their 1000th terms?
SOL:-
For AP(1) a100 =a+(100-1)d
a100 = a +99d
AP(2) A100 =A +(100-1)d
=A+99d
AP(1) a100 -AP(2)A100=100
a +99d -A-99d =100
a -A =100
AP(1) a1000 =a+(1000-1)d = a +999d
AP(2)A1000 =A +(1000-1)d = A +999d
AP(1) a1000- AP(2)A1000 = a +999d- A -999d =a -A
a -A =100
so diffrence between APs 1000th term is also 100.
QUE:-13. How many three-digit numbers are divisible by 7?
SOL:-
3 digit numbers divisible by 7 ; 105 ,112,119, …………..,994
a(n) =a+(n-1)d
994 = 105 +(n-1)7
994 =105 +7n -7
994 = 98+7n
7n =994-98 =896
n =896/7=128
3 digit numbers divisible by 7 is 128 .
QUE:-14. How many multiples of 4 lie between 10 and 250?
SOL:-
multiples of 4 lie between 10 and 250 ; 12,16,20,24,……………248
a =12 ,d =4 n =?
a(n) = a +(n-1)d
a(n) = 12 +(n-1)4
248 = 12 +4n -4
4n = 248-8=240
n =240/4=60
multiples of 4 lie between 10 and 250 is 60.
QUE:-15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
SOL:-
For first AP A =63 , D = 65-63=2 ,A(n) =?
A(n) =A +(n-1)D
A(n) = 63 +(n-1)2 = 63 +2n -2 = 61 +2n
For second AP a= 3 ,d=10-3 =3 ,a(n) =?
a(n) = a +(n-1)d
a(n) =3 + (n-1)7 = 3 +7n -7 =-4 +7n
according to the question
A(n) = a(n)
61 +2n =-4 +7n
2n -7n = -4 -61
-5n =-65
n =-65/-5=13
so the 13th term of APs are equal .
QUE:-16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
SOL:-
let the first term of AP is a
and common diffrence is d
according to the question
3rd term of AP =16
a +2d =16 ………….(1)
and a+ 6d =a+4d +12
a +6d -a -4d =12
2d =12
d =12/2 =6
putting the value of d in equ. (1)
a +2×6 =16
a =16-12 =4
so AP is 4 , 10 ,16,22, 28,34,40, ………….
5th term 28 and 7th term is 40
7th term – 5th term =12
40 -28 =12
QUE:-17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
SOL:-
a =3 ,d=8-3=5
253 = 3+(n-1)5
253 = 3 +5n -5
5n =253 +2
n = 255/5
n = 51
20th term from last 51 -19 =32
a31 =3 +(32-1)5= 3+155 =158
QUE:-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.
SOL:-
sum of 4th and 8th term of AP =24
a(4) = a +(4-1)d =a +3d
a(8) =a + (8-1)d =a + 7d
a(4) + a(8) =24
a +3d +a + 7d =24
2a +10d =24
a +5d =12 ………(1)
a(10) =a +(10-1)d =a +9d =44
a =44 -9d ………..(2)
putting the value of a in equ (1)
44-9d+5d = 12
-4d =12-44
d =-32/-4=8
putting the value of d in equ. (2)
a =44 -9×8 = -28
a = -28
AP is -28 , -20 ,-12,-4 ,
QUE:-19 Subba Rao started work in 1995 at an annual salary of ` 5000 and received an increment
of ` 200 each year. In which year did his income reach ` 7000?
SOL:-
the salary of Subbo Rao is a = 5000
increment per year is d =200
a(n) =7000
n =?
a(n) = a +(n-1)d
7000 = 5000 +(n -1)200
7000 = 5000 +200n -200
200n = 7000-4800
n = 2200/200=11 th year
so in year 1995+11 =2006 his income reach 7000.
QUE:-20 Ramkali saved ` 5 in the first week of a year and then increased her weekly savings by
` 1.75. If in the nth week, her weekly savings become ` 20.75, find n.
SOL:-
Ramkali save in first week is a =5
increased saving weekly d =1.75
weekly saving become a(n) =20.75
nth week n =?
a(n) = a + (n-1)d
20.75 = 5 + (n-1)1.75
20.75 = 5+1.75 n-1.75
20.75 = 3.25 +1.75n
1.75n =20.75-3.25=17.50
n =17.50/1.75=10
so in the 10th week her saving become 20.75.
You can see the solution for complete chapter here –
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