Question

Find A. P whose fourth term is 9 and the sum is of its sixth term and thirteenth term is 40.

Answer 1

heya mate !!

here's your answer :-

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Given that ,

● The fourth term of an A.P. is 9.

According to the question ,

An = a + ( n - 1 ) d

===> a + ( 5 - 1 ) d = 9

===> a + 3d = 9 ............( 1 )

● The sum of it 6th term and 13th term is 40.

According to the question ,

6th term + 13th term = 40

[ a + ( n - 1 ) d ] + [ a + ( n - 1 ) d ] = 40

[ a + ( 6 - 1 ) d ] + [ a + ( 13 - 1 ) d ] = 40

[ a + 5d ] + [ a + 12d ] = 40

2a + 17d = 40 ......................( 2 )

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● Subtract ( 2 ) from ( 1 )

We have to multiply by 2 first supposition to make it subtracted.

==> a + 3d = 9 multiply by 2 we get

==> 2a + 6d = 18
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On subtraction , we get

==> 2a + 17d = 40

==> 2a + 6d = 18
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==> + 11d = 22

==> 11d = 22

==> d = 2

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Put the value of d in eq. ( 1 ) we get ,

==> 2a + 6d = 18

==> 2a + 6 * 2 = 18

==> 2a + 12 = 18

==> 2a = 6

==> a = 3

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hence we get , a = 3 and d = 2

So , arithmetic progress =

☆ a , a + d , a + 2d , a + 3d , a + 4d , a + 5d , a + 6d , a + 7d , a + 8d , a + 9d , a + 10d , a + 11d + a + 12d.

On putting values , we get

3 , 5 ,7 ,9 , 11 , 13 ,15 , 17 , 19 , 21 , 23 , 25 , 27 and so on.
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hope it helps !!

thanks for asking :)

☆ be brainly ☆