Find A. P whose fourth term is 9 and the sum is of its sixth term and thirteenth term is 40.
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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.
__________________________
hope it helps !!
thanks for asking :)
☆ be brainly ☆
here's your answer :-
__________________________
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 )
__________________________
● 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
__________________________
On subtraction , we get
==> 2a + 17d = 40
==> 2a + 6d = 18
______________
==> + 11d = 22
==> 11d = 22
==> d = 2
________________________
Put the value of d in eq. ( 1 ) we get ,
==> 2a + 6d = 18
==> 2a + 6 * 2 = 18
==> 2a + 12 = 18
==> 2a = 6
==> a = 3
__________________________
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.
__________________________
hope it helps !!
thanks for asking :)
☆ be brainly ☆
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