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Sequences and Series Sequence and Series

Find the sum to indicated number of terms in each of the geometric progressions inExercises 39 to 42:

39. 0.15, 0.015, 0.0015, . 20 terms.

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Payal Gupta

Contributor-Level 10

39. Here a = 0.15 = $\frac{15}{100}$

$r = \frac{0.015}{0.15} = \frac{\frac{15}{1000}}{\frac{15}{150}} = \frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_{n} = \frac{a (1 - r^{n})}{1 - r}$

$\Rightarrow$ $s_{20} = \frac{\frac{15}{100} [1 - {(\frac{1}{10})}^{20}]}{1 - \frac{1}{10}}$

= $\frac{\frac{15}{100} [1 - {(0.1)}^{20}]}{\frac{9}{10}}$

= $\frac{15}{100} * \frac{10}{9} [1 - {(0.1)}^{20}]$

= $\frac{1}{6} [1 - {(0.1)}^{20}]$

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Sequences and Series Sequence and Series

38. For what values of x, the numbers $- \frac{2}{7}$ , $x$ , $- \frac{7}{2}$ are in G.P.?

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Payal Gupta

Contributor-Level 10

38. For $\frac{- 2}{7}$ , $x$ , $\frac{- 7}{2}$ to be in G.P. we have the condition,

$\Rightarrow$ $\frac{x}{- 2 / 7} = \frac{- 7 / 2}{x}$

$\Rightarrow$ $x^{2} = (\frac{- 7}{2}) * (\frac{- 2}{7})$

$\Rightarrow$ $x^{2} = 1$

$\Rightarrow$ $x = + 1$

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Sequences and Series Sequence and Series

37. Which term of the following sequences:

(a) 2,2√2 ,4,. is 128? (b) √3,3,3√3 ,. is729 ?

(c) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots i s \frac{1}{19683}$

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Sequences and Series Sequence and Series

36. The 4th term of a G.P. is square of its second term, and the first term is – 3.Determine its 7th term.

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Payal Gupta

Contributor-Level 10

36. Given, a = 3

Let r be the common ratio of the G.P.

Then, a₄= (a₂)²

$\Rightarrow$ ar^4-1 = (ar²^-1)²

$\Rightarrow$ ar³= (ar)²

$\Rightarrow$ ar³= a²r²

$\Rightarrow$ $\frac{r^{3}}{r^{2}} = \frac{a^{2}}{2}$

$\Rightarrow$ r = a = 3

$∴$ a₇ = ar⁷^-1 = ar⁶= (-3) (-3)⁶ = (-3)⁷= -2187.

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Sequences and Series Sequence and Series

35. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Showthat q² = ps.

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Payal Gupta

Contributor-Level 10

35. Let a and r be the first term and the common ratio between G.P.

So, a₅= p $\Rightarrow$ a r^t^-1 = p $\Rightarrow$ ar⁴ = p

a₈ = q $\Rightarrow$ a r^8-1 =q $\Rightarrow$ ar⁷= q

a₁₁= r $\Rightarrow$ ar^11-1 = r $\Rightarrow$ ar¹⁰= 5

So, L.H.S. q^{2 =} (ar⁷)² = a²r¹⁴

R.H. S. = p.s= (ar⁴) (ar)¹⁰= a¹⁺¹r⁴⁺¹⁰ = a²r¹⁴

$∴$ L.H.S. = R.H.S.

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Sequences and Series Sequence and Series

34. Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

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Payal Gupta

Contributor-Level 10

34. Given, r=2

Let a be the first term,

Then, a₈= 92

$\Rightarrow$ ar⁸^-1=192

$\Rightarrow$ a (2)⁷ = 192

$\Rightarrow$ a = $\frac{192}{2^{7}} = \frac{192}{128} = \frac{3}{2}$

so, a = $\frac{3}{\begin{array}{l} 2 \end{array}}$

$∴$ a₁₂ = ar^12-1= $\begin{array}{l} (\frac{3}{2}) \end{array}$ (2)¹¹ = 3 2¹¹^-1

= 3*2¹⁰

= 31024

= 3072.

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Sequences and Series Sequence and Series

33. Find the 20th and nth terms of the G.P. $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, . . .$

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Payal Gupta

Contributor-Level 10

33. Here, a= $\frac{5}{2}$

= $r = \frac{\frac{5}{4}}{\frac{5}{2}} = \frac{1}{2}$

so, a_n = arⁿ^-1

i.e a₂₀=ar²⁰^-1= $(\frac{5}{2}) * {(\frac{1}{2})}^{19} = \frac{5}{2} * \frac{1}{2^{19}} = \frac{5}{2^{19 + 1}} = \frac{5}{2^{20}}$

and an= arⁿ^-¹= $(\frac{5}{2}) * {(\frac{1}{2})}^{n - 1} = \frac{5}{2} * \frac{1}{2^{n - 1}} = \frac{5}{2^{n - 1 + 1}} = \frac{5}{2^{n}}$

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Sequences and Series Sequence and Series

32. The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120° , find the number of the sides of the polygon.

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Payal Gupta

Contributor-Level 10

32. Let 'n' be the no. of sides of a polygon.

So, sum of all angles of a polygon with sides n

= (2n – 4) * 90°

= (n – 2) * 2 * 90°

= (n – 2)180°

As the smallest angle is 120° and the difference between 2 consecutive interior angle is 5°. We have,

a =120°

d =5°

So, sum of n sides =180° (n – 2).

$\frac{n}{2} [2 a + (n - 1) d] = 18 0^{°} (n - 2)$

$\frac{n}{2} [2 * 12 0^{°} + (n - 1) 5^{°}] = 18 0^{°} (n - 2)$

n [240°+5°n – 5°]=360°n– 720°

n [5°n+235°]=360°n – 720°

5°n²+235°n=360°n – 720°

n²+47°n=72n – 144 [? dividing by 5]

n²+47n – 72n+144=0.

n² – 25n+144=0

n² – 9n– 16n+144=0

n (n – 9) –16 (n – 9)=0

(n – 9) (n – 16)=0

So, n=9,16.

When n=9,

the largest angle =a+ (9 – 1)d=120°+8 * 5° = 120° + 40° = 16

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32. Let 'n' be the no. of sides of a polygon.

So, sum of all angles of a polygon with sides n

= (2n – 4) * 90°

= (n – 2) * 2 * 90°

= (n – 2)180°

As the smallest angle is 120° and the difference between 2 consecutive interior angle is 5°. We have,

a =120°

d =5°

So, sum of n sides =180° (n – 2).

$\frac{n}{2} [2 a + (n - 1) d] = 18 0^{°} (n - 2)$

$\frac{n}{2} [2 * 12 0^{°} + (n - 1) 5^{°}] = 18 0^{°} (n - 2)$

n [240°+5°n – 5°]=360°n– 720°

n [5°n+235°]=360°n – 720°

5°n²+235°n=360°n – 720°

n²+47°n=72n – 144 [? dividing by 5]

n²+47n – 72n+144=0.

n² – 25n+144=0

n² – 9n– 16n+144=0

n (n – 9) –16 (n – 9)=0

(n – 9) (n – 16)=0

So, n=9,16.

When n=9,

the largest angle =a+ (9 – 1)d=120°+8 * 5° = 120° + 40° = 160°< 180°

Hence, n=9 is permissible

When n=16,

Lorgest angle =a+ (16 – 1)d=120°+15 * 5° = 120° + 75° = 195°> 180°.

which is not permissible as no angle in polygon can exceed 180°.

Hence, number of sides in polygon, n=9.

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Sequences and Series Sequence and Series

31. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30th instalment?

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Payal Gupta

Contributor-Level 10

31. Since the man starts paying installmentas? 100 and increases? 5 every month.

a=100

d=5.

So, the A.P is 100,105,110,115, ….

? Amount of 30thinstallment=30th term of A.P =a₃₀

=a+ (30 – 1)d

=100+29 * 5.

=100+145

= ?245

i.e., He will pay? 245 in his 30th instalment.

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Sequences and Series Sequence and Series

30. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.

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Payal Gupta

Contributor-Level 10

30. Let A₁, A₂, A₃ . A_m be the m terms such that,

1, A₁, A₂, A₃, . A_m, 31 is an A.P.

So, a=1, first term of A.P

n=m+2, no. of term of A.P

l=31, last term of A.P. or (m+2)th term.

a+ [ (m+2) –1]d =31

1 + [m+1]d =31

(m+1)d =31 – 1=30

d = $\frac{30}{m + 1}$

The ratio of 7th and (m – 1) number is

$\frac{a + 7 d}{a + (m - 1) d} = \frac{5}{9}$ ? a₁term=a

a₂ =A1=a+d

a₃ =A2=a+2d

a₈ = A₇ = a + 7d

9 [a+7d]=5 [a+ (m – 1)d]

9a+63d=5a+5 (m – 1)d.

9a – 5a=5 (m – 1)d – 63d.

4a= [5 (m – 1) –63]d.

So, putting a=1 and d= $\frac{30}{m + 1}$

4 * 1= [5 (m – 1) –63] * $\frac{30}{m + 1}$

4 (m+1)= [5 (m – 1) –63] * 30

4m+4= [5m – 5 – 63] * 30

4m+4 =150m – 68 * 30 => 150m – 2040

150m – 4m =2040+4

146m =2044.

m = $\frac{2044}{146}$

m =

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30. Let A₁, A₂, A₃ . A_m be the m terms such that,

1, A₁, A₂, A₃, . A_m, 31 is an A.P.

So, a=1, first term of A.P

n=m+2, no. of term of A.P

l=31, last term of A.P. or (m+2)th term.

a+ [ (m+2) –1]d =31

1 + [m+1]d =31

(m+1)d =31 – 1=30

d = $\frac{30}{m + 1}$

The ratio of 7th and (m – 1) number is

$\frac{a + 7 d}{a + (m - 1) d} = \frac{5}{9}$ ? a₁term=a

a₂ =A1=a+d

a₃ =A2=a+2d

a₈ = A₇ = a + 7d

9 [a+7d]=5 [a+ (m – 1)d]

9a+63d=5a+5 (m – 1)d.

9a – 5a=5 (m – 1)d – 63d.

4a= [5 (m – 1) –63]d.

So, putting a=1 and d= $\frac{30}{m + 1}$

4 * 1= [5 (m – 1) –63] * $\frac{30}{m + 1}$

4 (m+1)= [5 (m – 1) –63] * 30

4m+4= [5m – 5 – 63] * 30

4m+4 =150m – 68 * 30 => 150m – 2040

150m – 4m =2040+4

146m =2044.

m = $\frac{2044}{146}$

m =14.

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