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Arithmetic Progressions Ex-5.3 Interview Questions Answers

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 = a–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      
(i)

First term, a = 7

nth term, a=84

Let 84 be the nth termof this A.P., then as per the nth term formula,

a= 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,

ana+(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 

Question 3 : In an AP

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 :