Question

How many terms of the A.P.
$20, \: \: 19\dfrac{1}{3} , \: 18 \dfrac{2}{3} ... \text{must \: be \: taken \: so \: that \: their \: sum \: is \: 300?}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given AP series is

$\rm :\longmapsto\:20, \: 19\dfrac{1}{3}, \: 18 \dfrac{2}{3}, \: - - -$

It means,

$\rm :\longmapsto\:First \: term \: of \: an \: AP \: series, a = 20$

$\rm :\longmapsto\:Common \: Difference, d = 19 \dfrac{1}{3} - 20$

$\rm \: \: = \: \dfrac{58}{3} - 20$

$\rm \: \: = \: \dfrac{58 - 20}{3}$

$\rm \: \: = \: \dfrac{ - 2}{3}$

$\rm :\longmapsto\:Common \: Difference, \: d = - \: \dfrac{2}{3}$

↝ Let number of terms be n.

According to statement,

↝ Sum of n terms of AP series, = 300

We know that,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

↝ On substituting the values, we get

$\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(2 \times 20 - \dfrac{2}{3} (n - 1) \bigg)$

$\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(40 - \dfrac{2}{3} (n - 1) \bigg)$

$\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(\dfrac{120 - 2n + 2}{3} \bigg)$

$\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(\dfrac{122 - 2n }{3} \bigg)$

$\rm :\longmapsto\:300 =n\bigg(\dfrac{61 - n }{3} \bigg)$

$\rm :\longmapsto\:900 = 61n - {n}^{2}$

$\rm :\longmapsto\: {n}^{2} - 61n + 900 = 0$

$\rm :\longmapsto\: {n}^{2} - 36n - 25n + 900 = 0$

$\rm :\longmapsto\:n(n - 36) - 25(n - 36) = 0$

$\rm :\longmapsto\:(n - 25)(n - 36) = 0$

$\bf\implies \:n = 25 \: \: \: or \: \: \: n = 36$

Justification of double answer,

Note :-

Here, double answer indicates that sum of 26th terms to 36th terms are 0.

Number of terms from 26th term to 36th term = 11

$\rm :\longmapsto\:d = - \: \dfrac{2}{3}$

$\rm :\longmapsto\:a_{26} = a + 25d = 20 - \dfrac{50}{3} = \dfrac{10}{3}$

Hence,

$\rm :\longmapsto\:S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg(2 \: \times \dfrac{10}{3} \: - \dfrac{2}{3} \:(11\:-\:1) \bigg)$

$\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg( \dfrac{20}{3} \: - \dfrac{2}{3} \:(10) \bigg)$

$\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg( \dfrac{20}{3} \: - \dfrac{20}{3} \bigg)$

$\rm :\longmapsto\:S_{11}\:=0$

Hence, justify

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Answer 2

Answer:

\large\underline{\sf{Solution-}}

Solution−

Given AP series is

\rm :\longmapsto\:20, \: 19\dfrac{1}{3}, \: 18 \dfrac{2}{3}, \: - - - :⟼20,19

3

1

,18

3

2

,−−−

It means,

\rm :\longmapsto\:First \: term \: of \: an \: AP \: series, a = 20:⟼FirsttermofanAPseries,a=20

\rm :\longmapsto\:Common \: Difference, d = 19 \dfrac{1}{3} - 20:⟼CommonDifference,d=19

3

1

−20

\rm \: \: = \: \dfrac{58}{3} - 20=

3

58

−20

\rm \: \: = \: \dfrac{58 - 20}{3} =

3

58−20

\rm \: \: = \: \dfrac{ - 2}{3} =

3

−2

\rm :\longmapsto\:Common \: Difference, \: d = - \: \dfrac{2}{3} :⟼CommonDifference,d=−

3

2

↝ Let number of terms be n.

According to statement,

↝ Sum of n terms of AP series, = 300

We know that,

↝ Sum of n terms of an arithmetic sequence is,

\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}\end{gathered}

★

S

n

=

2

n

(2a+(n−1)d)

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

↝ On substituting the values, we get

\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(2 \times 20 - \dfrac{2}{3} (n - 1) \bigg) :⟼300=

2

n

(2×20−

3

2

(n−1))

\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(40 - \dfrac{2}{3} (n - 1) \bigg) :⟼300=

2

n

(40−

3

2

(n−1))

\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(\dfrac{120 - 2n + 2}{3} \bigg) :⟼300=

2

n

(

3

120−2n+2

)

\rm :\longmapsto\:300 = \dfrac{n}{2}\bigg(\dfrac{122 - 2n }{3} \bigg) :⟼300=

2

n

(

3

122−2n

)

\rm :\longmapsto\:300 =n\bigg(\dfrac{61 - n }{3} \bigg) :⟼300=n(

3

61−n

)

\rm :\longmapsto\:900 = 61n - {n}^{2} :⟼900=61n−n

2

\rm :\longmapsto\: {n}^{2} - 61n + 900 = 0:⟼n

2

−61n+900=0

\rm :\longmapsto\: {n}^{2} - 36n - 25n + 900 = 0:⟼n

2

−36n−25n+900=0

\rm :\longmapsto\:n(n - 36) - 25(n - 36) = 0:⟼n(n−36)−25(n−36)=0

\rm :\longmapsto\:(n - 25)(n - 36) = 0:⟼(n−25)(n−36)=0

\bf\implies \:n = 25 \: \: \: or \: \: \: n = 36⟹n=25orn=36

Justification of double answer,

Note :-

Here, double answer indicates that sum of 26th terms to 36th terms are 0.

Number of terms from 26th term to 36th term = 11

\rm :\longmapsto\:d = - \: \dfrac{2}{3} :⟼d=−

3

2

\rm :\longmapsto\:a_{26} = a + 25d = 20 - \dfrac{50}{3} = \dfrac{10}{3} :⟼a

26

=a+25d=20−

3

50

=

3

10

Hence,

\rm :\longmapsto\:S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg):⟼S

n

=

2

n

(2a+(n−1)d)

\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg(2 \: \times \dfrac{10}{3} \: - \dfrac{2}{3} \:(11\:-\:1) \bigg):⟼S

11

=

2

11

(2×

3

10

−

3

2

(11−1))

\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg( \dfrac{20}{3} \: - \dfrac{2}{3} \:(10) \bigg):⟼S

11

=

2

11

(

3

20

−

3

2

(10))

\rm :\longmapsto\:S_{11}\:=\dfrac{11}{2} \bigg( \dfrac{20}{3} \: - \dfrac{20}{3} \bigg):⟼S

11

=

2

11

(

3

20

−

3

20

)

\rm :\longmapsto\:S_{11}\:=0:⟼S

11

=0

Hence, justify

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

\begin{gathered}\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}\end{gathered}

★

a

n

=a+(n−1)d

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Step-by-step explanation:

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Warm regards:Miss Chikchiki