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

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