class 10th maths chapter5 ....5.3 question 3 all parts ....plzz solive all parts .....it,s urgent in english not in hindi plzz
Answers
Step-by-step explanation:
Solutions:
(i) Given, 2, 7, 12 ,…, to 10 terms
For this A.P.,
first term, a = 2
And common difference, d = a2 − a1 = 7−2 = 5
n = 10
We know that, the formula for sum of nth term in AP series is,
Sn = n/2 [2a +(n-1)d]
S10 = 10/2 [2(2)+(10 -1)×5]
= 5[4+(9)×(5)]
= 5 × 49 = 245
(ii) Given, −37, −33, −29 ,…, to 12 terms
For this A.P.,
first term, a = −37
And common difference, d = a2− a1
d= (−33)−(−37)
= − 33 + 37 = 4
n = 12
We know that, the formula for sum of nth term in AP series is,
Sn = n/2 [2a+(n-1)d]
S12 = 12/2 [2(-37)+(12-1)×4]
= 6[-74+11×4]
= 6[-74+44]
= 6(-30) = -180
(iii) Given, 0.6, 1.7, 2.8 ,…, to 100 terms
For this A.P.,
first term, a = 0.6
Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that, the formula for sum of nth term in AP series is,
Sn = n/2[2a +(n-1)d]
S12 = 50/2 [1.2+(99)×1.1]
= 50[1.2+108.9]
= 50[110.1]
= 5505
(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms
For this A.P.,
First term, a = 1/5
Common difference, d = a2 –a1 = (1/12)-(1/5) = 1/60
And number of terms n = 11
We know that, the formula for sum of nth term in AP series is,
Sn = n/2 [2a + (n – 1) d]
Sn = \frac{11}{2}\left [ 2\left ( \frac{1}{15} \right ) + \frac{11-1}{60}\right ]
2
11
[2(
15
1
)+
60
11−1
]
= 11/2(2/15 + 10/60)
= 11/2 (9/30)
= 33/20
2.
(i) 7 + 10\frac{1}{2}10
2
1
+ 14 + ….. + 84
(ii) 34 + 32 + 30 + ……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Solutions:
(i) 7 + 10\frac{1}{2}10
2
1
+ 14 + ….. + 84
First term, a = 7
nth term, an = 84,
Common difference, d = a2 – a1
= 10\frac{1}{2}10
2
1
– 7 = \frac{21}{2} – 7
2
21
–7 = \frac{7}{2}
2
7
Let 84 be the nth term of this A.P., then as per the nth term formula,
an = a(n-1)d
84 = 7+(n – 1)×7/2
77 = (n-1)×7/2
22 = n−1
n = 23
We know that, sum of n term is;
Sn = n/2 (a + l) , l = 84
Sn = 23/2 (7+84)
Sn = (23×91/2) = 2093/2
Sn = 1046\frac{1}{2}
2
1
(ii) Given, 34 + 32 + 30 + ……….. + 10
For this A.P.,
first term, a = 34
common difference, d = a2−a1 = 32−34 = −2
nth term, an= 10
Let 10 be the nth term of this A.P., therefore,
an= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
We know that, sum of n terms is;
Sn = n/2 (a +l) , l = 10
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
(iii) Given, (−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
First term, a = −5
nth term, an= −230
Common difference, d = a2−a1 = (−8)−(−5)
⇒d = − 8+5 = −3
Let −230 be the nth term of this A.P., and by the nth term formula we know,
an= a+(n−1)d
−230 = − 5+(n−1)(−3)
−225 = (n−1)(−3)
(n−1) = 75
n = 76
And, Sum of n term,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930
Solutions:
Let there be n terms of the AP. 9, 17, 25 …
For this A.P.,
First term, a = 9
Common difference, d = a2−a1 = 17−9 = 8
As, the sum of n terms, is;
Sn = n/2 [2a+(n -1)d]
636 = n/2 [2×a+(8-1)×8]
636 = n/2 [18+(n-1)×8]
636 = n [9 +4n −4]
636 = n (4n +5)
4n2 +5n −636 = 0
4n2 +53n −48n −636 = 0
n (4n + 53)−12 (4n + 53) = 0
(4n +53)(n −12) = 0
Either 4n+53 = 0 or n−12 = 0
n = (-53/4) or n = 12
n cannot be negative or fraction, therefore, n = 12 only.
5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:Given that,
first term, a = 5
last term, l = 45
Sum of the AP, Sn = 400
As we know, the sum of AP formula is;
Sn = n/2 (a+l)
400 = n/2(5+45)
400 = n/2(50)
Number of terms, n =16
As we know, the last term of AP series can be written as;
l = a+(n −1)d
45 = 5 +(16 −1)d
40 = 15d
Common difference, d = 40/15 = 8/3
6.
Solution: Given that,
First term, a = 17
Last term, l = 350
Common difference, d = 9
Let there be n terms in the A.P., thus the formula for last term can be written as;
l = a+(n −1)d
350 = 17+(n −1)9
333 = (n−1)9
(n−1) = 37
n = 38
Sn = n/2 (a+l)
S38 = 13/2 (17+350)
= 19×367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
7.
Solution:Given,
Common difference, d = 7
22nd term, a22 = 149
Sum of first 22 term, S22 = ?
By the formula of nth term,
an = a+(n−1)d
a22 = a+(22−1)d
149 = a+21×7
149 = a+147
a = 2 = First term
Sum of n terms,
Sn = n/2(a+an)
= 22/2 (2+149)
= 11×151
= 1661
u can also search on Google
because most of ans notvsent
Answer:
hey dear,
plz see my answers to your 3rd questions
if you are going to forgive me ,plz thanks my answers
With Regards ,
Nandlal Singh