Question

The sum of first 7 term of AP is 63 and sum of is next 7 terms IE 161 find the 28th term of an AP

Answer 1

☜☆☞hey friend!!! ☜☆☞

here is your answer ☞
→_→→_→→_→→_→→_→

Sum of the first n terms of an A.P, Sn = ( n / 2) [ 2a + ( n -1)d ]

Given that sum of the first 7 terms of an A.P is 63 i. e S7 = 63.

⇒ ( 7 / 2) [ 2a + 6d ] = 63

⇒ 2a + 6d = 18 --------(1)

Also given sum of its next 7 terms is 161.

But Sum of first 14 terms = sum of first 7 terms + sum of next 7 terms.

S14 = 63 + 161 = 224

⇒ ( 14 / 2) [ 2a + 13d ] = 224.

⇒ 7 [ 2a + 13d ] = 224.

⇒ [ 2a + 13d ] = 32 -------92)

Solving equ (1) and (2) we obtain

d = 2 and a = 3.

Now t28 = a + ( 28 - 1) d

t28 = 3 + ( 28 - 1) 2

t28 = 57.

∴28th term of this A.P. is 57.

Hope it will help you

Devil_king ▄︻̷̿┻̿═━一

Answer 2

heya mate !!

here's your answer !!
_________________________

The sum of 7th term of an A.P. = 63

Sn = [ n/2 ] ( 2a + n - 1 ) d ]

63 = 7/2 ( 2a + 7 - 1 ) d

63 = 7/2 ( 2a + 6d )

( 63 × 2 ) / 7 = 2a + 6d

18 = 2a + 6d ............( 1 )
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The sum of next 7th is = 161

=> sum of seven term + sum of next seven term

=> The sum of 14th term = 161 + 63

=> 224

So , the sum of 14th term = [ n / 2 ( 2a + ( n - 1 ) d

=> 14/2 ( 2a + 13d ) = 224

=> 7 ( 2a + 13 d ) = 224

=> 2a + 13d = 224 .............( 2 )
__________________________

On subtracting ( 2 ) from ( 1 ) , we get

=> ( 2a + 13d = 32 ) - ( 2a + 6d ) = 18

=> 7d = 14

=> d = 2

Putting the value d in eq. ( 2 ) , we get

2a + 6d = 18

2a + 6 ( 2 ) = 18

2a + 12 = 18

2a = 6

a = 3

__________________________

So , 28th term of an A.P. will be :-

=> a + ( n - 1 ) d

=> 3 + ( 28 - 1 ) 2

=> 3 + 54

=> 57

hence , 28 term of an A.P. is 57

_______________________

hope it helps !!

thanks for asking !!

☆ be brainly ☆