Arithmetic Progressions Ex 5.3 Solutions
Question - 1 : - Find the sum of the following APs.
Answer - 1 : -
(i) 2, 7, 12 ,…., to10 terms.
(ii) − 37, − 33, − 29 ,…, to 12 terms
(iii) 0.6, 1.7, 2.8 ,…….., to 100 terms
(iv) 1/15, 1/12, 1/10, …… , to 11 terms
Solution
(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 termin 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
Solution
(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 termin 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
Solution
(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 termin 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
Solution
(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 termin AP series is,
Sn = n/2 [2a + (n –1) d]
= 11/2(2/15 + 10/60)
= 11/2 (9/30)
= 33/20
Question - 2 : - Find the sums givenbelow:
Answer - 2 : - 
(ii) 34 + 32 + 30 +……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Solution 
First term, a = 7
nth term, an =84
Let 84 be the nth termof 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
Solution
(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 termof 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
Solution
(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 termof 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
Answer - 3 : -
(i) Given a =5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35,find d and S13.
(iii) Given a12 = 37, d = 3,find a and S12.
(iv) Given a3 = 15, S10 =125, find d and a10.
(v) Given d = 5, S9 = 75,find a and a9.
(vi) Given a = 2, d = 8, Sn =90, find n and an.
(vii) Given a = 8, an = 62, Sn =210, find n and d.
(viii) Given an = 4, d = 2, Sn =− 14, find n and a.
(ix) Given a = 3, n = 8, S =192, find d.
(x) Given l = 28, S = 144 and there are total9 terms. Find a.
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Question - 4 : - How many terms of AP: 9, 17, 25, … must be taken to give a sum of 636?
Answer - 4 : - 
Question - 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.
Answer - 5 : - 
Question - 6 : - The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer - 6 : - 
Question - 7 : - Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer - 7 : - 
Question - 8 : - Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer - 8 : - 
Question - 9 : - If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer - 9 : - 
Question - 10 : - Show that a1, a2, ……. an,…… form an AP where an is defined as below:
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case
Answer - 10 : -
Solution:
