If the 9th term is 499 and the 499th term is 9 then sum of first 508 terms
Answers
Answer:
Hi Mate !!
a ( First term )
d ( Common difference )
n ( no. of term )
an ( nth term )
• an = a + ( n - 1 )d
• Given :- 9th term is 499 i.e ,
a9 = 499 ..... ( When n = 9 ) ..... ( i )
499th term is 9 i.e ,
a499 = 9 ..... ( When n = 499 ) .... ( ii )
From ( i )
a9 = 499
a + 8d = 499 .... ( iii )
From ( ii )
a499 = 9
a + 498d = 9 ...... ( iv )
Subtracting ( iv ) from ( iii )
a + 8d - ( a + 498d ) = 499 - 9
a + 8d - a - 498d = 490
- 490d = 490
d = ( - 1 )
Putting value of d in eq ( iii )
a + 8d = 499
a + 8 × ( - 1 ) = 499
a - 8 = 499
a = 499 + 8
a = 507
Now , we have value of a and d , value of an is already given i.e, 0 ( Zero ) and we have to find value of n !!
As ,
an = a + ( n - 1 ) d
0 = 507 + ( n - 1 ) × ( - 1 )
0 = 507 - n + 1
0 = 508 - n
- 508 = - n
n = 508 !!
Hence , the value of an at n = 508 will be Zero !!
° Verification :-
an = a + ( n - 1 ) × d
an = 507 + ( 508 - 1 ) × ( - 1 )
an = 507 + 507 × ( - 1 )
an = 507 - 507
an = 0
Hence , verified !!
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