Math, asked by sangithakailash50371, 2 months ago

The sum of first 15th term of an arithmetic sequence is 570 and its 12th trem is 62
What is the sum of 30th term

Answers

Answered by mathdude500
3

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

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.

Now, Given that

Sum of first 15 terms of an AP = 570

\red{\rm :\longmapsto\:S_{15}\:=\dfrac{15}{2} \bigg(2 \:a\:+\:(15\:-\:1)\:d \bigg)}

\rm :\longmapsto\:570 = \dfrac{15}{2}(2a + 14d)

\rm :\longmapsto\:38 = \dfrac{2}{2}(a + 7d)

\bf\implies \:a + 7d = 38 -  -  - (1)

Also, we know that

↝ 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.

Now, Given that

\rm :\longmapsto\:a_{12}\:=62

\rm :\longmapsto\:a + (12 - 1)d = 62

\rm :\longmapsto\:a + 11d = 62 -  -  - (2)

On Subtracting equation (1) from equation (2), we get

\rm :\longmapsto\:4d = 24

\bf\implies \:d = 6

On substituting the value of d, in equation (1) we get

\rm :\longmapsto\:a + 7 \times 6 = 38

\rm :\longmapsto\:a + 42= 38

\rm :\longmapsto\:a = 38 - 42

\bf\implies \:a =  - 4

Now, Sum of first 30 terms is given by

\rm :\longmapsto\:S_{30}\:=\dfrac{30}{2} \bigg(2 \:( - 4)\:+\:(30\:-\:1)\:6 \bigg)

\rm :\longmapsto\:S_{30}\:=15 \bigg( -  \:8\:+\:(29)\:6 \bigg)

\rm :\longmapsto\:S_{30}\:=15 \bigg( -  \:8\:+\:174 \bigg)

\rm :\longmapsto\:S_{30}\:=15  \times 166

\bf :\longmapsto\:S_{30}\:=2490

Answered by XxItzAdyashaxX
2

Answer:

We know that

s

n

=

n

2

(

2

a

+

(

n

1

)

d

)

Thus

186

=

12

2

(

2

a

+

(

11

)

d

)

186

=

6

(

2

a

+

11

d

)

31

=

2

a

+

11

d

Now we know that

t

n

=

a

+

(

n

1

)

d

.

83

=

a

+

(

20

1

)

d

83

=

a

+

19

d

We now have a system of equations:

{

31

=

2

a

+

11

d

83

=

a

+

19

d

Substituting (2) into (1), we get

31

=

2

(

83

19

d

)

+

11

d

31

=

166

38

d

+

11

d

135

=

27

d

d

=

5

Now solving for

a

:

83

19

(

5

)

=

12

The sum is once again given by

s

40

=

40

2

(

2

(

12

)

+

(

39

)

5

)

s

40

=

20

(

171

)

s

40

=

3420

Hopefully this helps!

Adyasha here

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