Question

in an arithmetic progression of 50, terms the sum of first 10 terms is 210 and the sum of last 15 terms is 2565 find the arithmetic progression

Answer 1

hope it will help you mate

Answer 2

hey......
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Sum of the first n term given by

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

putting n = 10 , we get

S10 = 10 / 2 [ 2a + ( 10 - 1 ) d ]

210 = 5 ( 2a + 9d )

2a + 9d = 42 ..................( 1 )

sum of the last 15 term is 2565

sum of the first 50 term - sum of the first 35 term = 2565

S50 - S35 = 2565

50 / 2 [ 2a + ( 50 - 1 ) d ] - 35 / 2 [ 2a + ( 35 - 1 )d ] = 2565

25 ( 2a + 49d ) - 35 / 2 [ 2a + 34d ] = 2565

2 ( 2a + 49d ) - 7 / 2 ( 2a + 34d ) = 513

10a + 245d - 7a + 119d = 513

3a + 126d = 513

a + 42d = 171................ ( 2 )

Multiply the equation ( 2 ) with 2 , we get


2a + 84d = 342.............. ( 3 )


Subtracting ( 1 ) from ( 3 )

d = 4

Now , substituting the value of d in equation ( 1 )

2a + 9d = 42

2a + 9 × 4 = 42

2a = 42 - 36

2a = 6

a = 3

So , the required A.P is

3 , 7 , 11 , 15 , 19 , 23 , 27 , 31 , 35 , 39 , ...........


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i hope this helps u.
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