CBSE 10th Maths - arithmetic - progressions - Exercise 5.3

Solution:

(i) Given, 2, 7, 12 ,…, to 10 terms

For this A.P.,

first term, a = 2

And common difference, d = $a_2 − a_1$ = 7−2 = 5

n = 10

We know that, the formula for sum of $n^{th}$ term in AP series is,

$S_n = \dfrac{n}{2} [2a +(n-1)d]$

$S_{10} = \dfrac{10}{2} [2(2)+(10 -1)×5]$

= 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 = $a_2− a_1$

d= (−33)−(−37) = − 33 + 37 = 4

n = 12

We know that, the formula for sum of nth term in AP series is,

$S_n = \dfrac{n}{2} [2a +(n-1)d]$

$S_{12} = \dfrac{12}{2} [2(-37)+(12-1)×4]$

= 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 = $a_2 − a_1$ = 1.7 − 0.6 = 1.1

n = 100

We know that, the formula for sum of nth term in AP series is,

$S_n = \dfrac{n}{2} [2a +(n-1)d]$

$S_{12} = \dfrac{50}{2} [1.2+(99)×1.1]$

= 50[1.2+108.9]

= 50[110.1]

= 5505

(iv) Given, $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}$, …… , to 11 terms

For this A.P.,

First term, a = $\dfrac{1}{5}$

Common difference, d = $a_2 –a_1 = (\dfrac{1}{12})-(\dfrac{1}{5}) = \dfrac{1}{60}$

And number of terms n = 11

We know that, the formula for sum of nth term in AP series is,

$S_n = \dfrac{n}{2} [2a +(n-1)d]$

$S_n = \dfrac{11}{2} [2(\dfrac{1}{15}) + \dfrac{(11-1)1}{60}]$

= $\dfrac{11}{2}(\dfrac{2}{15} + \dfrac{10}{60})$

=$ \dfrac{11}{2} (\dfrac{9}{30})$

= $\dfrac{33}{20}$

Solution:

(i) given 7 +10 $\dfrac{1}{2}$ ,+ 14 + . . . + 84

First term, a = 7

$n^{th}$ term, $a_n$ = 84

Common difference, d = $a_2 –a_1 = 10 \dfrac{1}{2} - 7 = \dfrac{21}{2} -7 = \dfrac{7}{2}$

Let 84 be the $n^{th}$ term of this A.P., then as per the $n^{th}$ term formula,

$a_n = a+(n-1)d$

84 = 7+(n – 1)×$\dfrac{7}{2}$

77 = (n-1)×$\dfrac{7}{2}$

22 = n−1

n = 23

We know that, sum of n term is;

$S_n = \dfrac{n}{2} (a + l)$

l = 84

$S_n = \dfrac{23}{2} (7+84)$

$S_n = (23×\dfrac{91}{2}) = \dfrac{2093}{2}$

$S_n = 1046 \dfrac{1}{2}$

(ii) Given, 34 + 32 + 30 + ……….. + 10

For this A.P.

first term, a = 34

common difference, $d = a_2−a_1$ = 32−34 = −2

$n^{th}$ term, $a_n$ = 10

Let 10 be the $n^{th}$ term of this A.P., therefore,

$a_n = a+(n-1)d$

10 = 34+(n−1)(−2)

−24 = (n −1)(−2)

12 = n −1

n = 13

We know that, sum of n terms is;

$S_n = \dfrac{n}{2} (a + l)$

l = 10

= $\dfrac{13}{2}$ (34 + 10)

= (13×$\dfrac{44}{2}$) = 13 × 22

= 286

(iii) Given, (−5) + (−8) + (−11) + ………… + (−230)

For this A.P.,

First term, a = −5

$n^{th}$ term, $a_n$= −230

Common difference, $d = a_2−a_1$ = (−8)−(−5)

d = − 8+5 = −3

Let −230 be the $n^{th}$ term of this A.P., and by the $n^{th}$ term formula we know,

$a_n = a+(n-1)d$

−230 = − 5+(n−1)(−3)

−225 = (n−1)(−3)

(n−1) = 75

n = 76

And, Sum of n term,

$S_n = \dfrac{n}{2} (a + l)$

= $\dfrac{76}{2} [(-5) + (-230)]$

= 38(-235)

= -8930

Solution:

(i) Given that, a = 5, d = 3, $a_n$ = 50

As we know, from the formula of the nth term in an AP,

$a_n = a+(n-1)d$

Therefore, putting the given values, we get,

50 = 5+(n -1)×3

3(n -1) = 45

n -1 = 15

n = 16

Now, sum of n terms,

$S_n = \dfrac{n}{2} (a+a_n)$

$S_n = \dfrac{16}{2}$ (5 + 50) = 440

(ii) Given that, a = 7, $a_{13}$ = 35

As we know, from the formula of the nth term in an AP,

$a_n = a+(n-1)d$

Therefore, putting the given values, we get,

35 = 7+(13-1)d

12d = 28

d = $\dfrac{28}{12}$ = 2.33

Now,

$S_n = \dfrac{n}{2} (a+a_n)$

$S_{13} = \dfrac{13}{2}$ (7+35) = 273

(iii) Given that, $a_{12}$ = 37, d = 3

As we know, from the formula of the nth term in an AP,

$a_n = a+(n-1)d$

Therefore, putting the given values, we get,

$a_{12}$ = a+(12−1)3

37 = a+33

a = 4

Now, sum of nth term,

$S_n = \dfrac{n}{2} (a+a_n)$

$S_n = \dfrac{12}{2} (4+37)$

= 246

(iv) Given that, $a_3 = 15, S_{10} = 125$

As we know, from the formula of the nth term in an AP,

$a_n = a+(n-1)d$

Therefore, putting the given values, we get,

$a_3$ = a+(3−1)d

15 = a+2d ………………………….. (i)

Sum of the nth term,

$S_n = \dfrac{n}{2} [2a+(n -1)d]$

$S_{10} = \dfrac{10}{2} [2a+(10-1)d]$

125 = 5(2a+9d)

25 = 2a+9d ……………………….. (ii)

On multiplying equation (i) by (ii), we will get;

30 = 2a+4d ………………………………. (iii)

By subtracting equation (iii) from (ii), we get,

−5 = 5d

d = −1

From equation (i),

15 = a+2(−1)

15 = a−2

a = 17 = First term

$a_{10}$ = a+(10−1)d

$a_{10}$ = 17+(9)(−1)

$a_{10}$ = 17−9 = 8

(v) Given that, d = 5, $S_9$ = 75

As, sum of n terms in AP is,

$S_n = \dfrac{n}{2} [2a+(n -1)d]$

Therefore, the sum of first nine terms are;

$S_9 = \dfrac{9}{2} [2a +(9-1)5]$

25 = 3(a+20)

25 = 3a+60

3a = 25−60

a = $-\dfrac{35}{3}$

As we know, the nth term can be written as

$a_n = a+(n-1)d$

$a_9$ = a+(9−1)(5)

= $-\dfrac{35}{3}+8(5)$

= $-\dfrac{35}{3}+40$

=$ (\dfrac{-35+120}{3}) =\dfrac{85}{3}$

(vi) Given that, a = 2, d = 8, $S_n$ = 90

As, sum of n terms in an AP is,

$S_n = \dfrac{n}{2} [2a+(n -1)d]$

90 = $\dfrac{n}{2}$ [2a +(n -1)d]

180 = n(4+8n -8) = n(8n-4) = $8n^2$-4n

$8n^2$-4n –180 = 0

$2n^2$–n-45 = 0

$2n^2$-10n+9n-45 = 0

2n(n -5)+9(n -5) = 0

(n-5)(2n+9) = 0

So, n = 5 (as n only be a positive integer)

$ a_5$ = 8+5×4 = 34

(vii) Given that, a = 8, $a_n$ = 62, $S_n$ = 210

As, sum of n terms in an AP is,

$S_n = \dfrac{n}{2} (a+a_n)$

210 = $\dfrac{n}{2}$ (8 +62)

35n = 210

n = $\dfrac{210}{35}$ = 6

Now, 62 = 8+5d

5d = 62-8 = 54

d = $\dfrac{54}{5}$ = 10.8

(viii) Given that, nth term, an = 4, common difference, d = 2, sum of n terms, $S_n$ = −14.

As we know, from the formula of the nth term in an AP,

$a_n = a+(n-1)d$

Therefore, putting the given values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………………………………. (i)

As we know, the sum of n terms is;

$S_n = \dfrac{n}{2} (a+a_n)$

-14 = $\dfrac{n}{2} (a+4)$

−28 = n (a+4)

−28 = n (6 −2n +4) {From equation (i)}

−28 = n (− 2n +10)

−28 = − $2n^2$+10n

$2n^2$ −10n − 28 = 0

$n^2$ −5n −14 = 0

$n^2$ −7n+2n −14 = 0

n (n−7)+2(n −7) = 0

(n −7)(n +2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14

= −8

(ix) Given that, first term, a = 3,

Number of terms, n = 8

And sum of n terms, S = 192

As we know,

$S_n = \dfrac{n}{2} [2a+(n -1)d]$

192 = $\dfrac{8}{2}$ [2×3+(8 -1)d]

192 = 4[6 +7d]

48 = 6+7d

42 = 7d

d = 6

(x) Given that, l = 28,S = 144 and there are total of 9 terms.

Sum of n terms formula,

$S_n = \dfrac{n}{2} (a + l)$

144 = $\dfrac{9}{2}$(a+28)

(16)×(2) = a+28

32 = a+28

a = 4

Solution:

Let there be n terms of the AP. 9, 17, 25 …

For this A.P.,

First term, a = 9

Common difference, $d = a_2−a_1$ = 17−9 = 8

As, the sum of n terms, is;

$S_n = \dfrac{n}{2} [2a+(n - 1)d]$

636 = $\dfrac{n}{2}$ [2×a+(8-1)×8]

636 = $\dfrac{n}{2}$ [18+(n-1)×8]

636 = n [9 +4n −4]

636 = n (4n +5)

$4n^2$ +5n −636 = 0

$4n^2$ +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 = $(-\dfrac{53}{4})$ or n = 12

n cannot be negative or fraction, therefore, n = 12 only.

Solution:

Given that,

first term, a = 5

last term, l = 45

Sum of the AP, $S_n$ = 400

As we know, the sum of AP formula is;

$S_n = \dfrac{n}{2} (a+l)$

400 = $\dfrac{n}{2}$(5+45)

400 = $\dfrac{n}{2}$(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 = $\dfrac{40}{15} = \dfrac{8}{3}$

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

$S_n = \dfrac{n}{2} (a+l)$

$S_{38} = \dfrac{38}{2}$ (17+350)

= 19×367

= 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

Solution:

Given,

Common difference, d = 7

$22^{nd}$ term, $a_{22}$ = 149

Sum of first 22 term, $S_{22}$ = ?

By the formula of $n^{th}$ term,

$a_n = a+(n−1)d$

$a_{22}$ = a+(22−1)d

149 = a+21×7

149 = a+147

a = 2 = First term

Sum of n terms,

$S_n =\dfrac{ n}{2}(a+a_n)$

$S_{22} = \dfrac{22}{2}$ (2+149)

= 11×151

= 1661

Solution:

Given that,

Second term, $a_2$ = 14

Third term, $a_3$ = 18

Common difference,$d = a_3−a_2$ = 18−14 = 4

$a_2$ = a+d

14 = a+4

a = 10 = First term

Sum of n terms;

$S_n = \dfrac{n}{2} [2a + (n – 1)d]$

$S_{51} = \dfrac{51}{2}$ [2×10 + (51-1) 4]

=$\dfrac{51}{2}$ [20+(50)×4]

= 51 × $\dfrac{220}{2}$

= 51 × 110

= 5610

Solution:

Given that,

$S_7$ = 49

$S_{17}$ = 289

We know, Sum of n terms;

$S_n = \dfrac{n}{2} [2a + (n – 1)d]$

Therefore,

$S_7= \dfrac{7}{2}$ [2a +(n -1)d]

$S_7 = \dfrac{7}{2}$ [2a + (7 -1)d]

49 = $\dfrac{7}{2}$ [2a +6d]

7 = (a+3d)

a + 3d = 7 …………………………………. (i)

In the same way,

$S_{17} = \dfrac{17}{2}$ [2a+(17-1)d]

289 =$\dfrac{17}{2}$ (2a +16d)

17 = (a+8d)

a +8d = 17 ………………………………. (ii)

Subtracting equation (i) from equation (ii),

5d = 10

d = 2

From equation (i), we can write it as;

a+3(2) = 7

a+ 6 = 7

a = 1

Hence,

$S_n = \dfrac{n}{2}[2a+(n-1)d]$

= $\dfrac{n}{2}[2(1)+(n – 1)×2]$

= $\dfrac{n}{2}(2+2n-2)$

= $\dfrac{n}{2}(2n)$

= $n^2$

Solution:

(i) $a_n$ = 3+4n

$a_1$ = 3+4(1) = 7

$a_2$ = 3+4(2) = 3+8 = 11

$a_3$ = 3+4(3) = 3+12 = 15

$a_4$ = 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

$a_2 − a_1$ = 11−7 = 4

$a_3 − a_2$ = 15−11 = 4

$a_4 − a_3$ = 19−15 = 4

Hence, $a_k + 1 − a_k$ is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

Now, we know, the sum of nth term is;

$S_n = \dfrac{n}{2}[2a+(n -1)d]$

$S_{15} = \dfrac{15}{2}[2(7)+(15-1)×4]$

= $\dfrac{15}{2}$[(14)+56]

= $\dfrac{15}{2}(70)$

= 15×35

= 525

(ii) $a_n$ = 9−5n

$a_1$ = 9−5×1 = 9−5 = 4

$a_2$ = 9−5×2 = 9−10 = −1

$a_3$ = 9−5×3 = 9−15 = −6

$a_4$ = 9−5×4 = 9−20 = −11

We can see here, the common difference between the terms are;

$a_2 − a_1$ = −1−4 = −5

$a_3 − a_2$ = −6−(−1) = −5

$a_4 − a_3$ = −11−(−6) = −5

Hence, $a_k + 1 − a_k$ is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

Now, we know, the sum of nth term is;

$S_n = \dfrac{n}{2} [2a +(n-1)d]$

$S_{15} = \dfrac{15}{2}$[2(4) +(15 -1)(-5)]

= $\dfrac{15}{2}$[8 +14(-5)]

= $\dfrac{15}{2}$(8-70)

= $\dfrac{15}{2}$(-62)

= 15(-31)

= -465