solve the arithmetic progression
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Sn=n/2 [2a+(n-1)d] (Sn=1050)
S14=14/2 [2×10+(14-1)d]
1050=7 (20+13d)
1050=140+91d
910=91d
d=10
a20=a+19d
=10+19×10
=200
hence 20th term is 200.
S14=14/2 [2×10+(14-1)d]
1050=7 (20+13d)
1050=140+91d
910=91d
d=10
a20=a+19d
=10+19×10
=200
hence 20th term is 200.
Answered by
3
Hi ,
Let a , d are first term and common
difference of an A.P
Sum of ' n ' terms = Sn
Sn = n/2 [ 2a + ( n - 1 )d ]
According to the problem given ,
S14 = 1050 , a = 10
14/2[ 2× 10 + ( 14 - 1 ) d ] = 1050
20 + 13d = 1050/7
13d = 150 - 20
13d = 130
d = 130/13
d = 10
Therefore ,
20th term in A.P = a + 19d
t20 = 10 + 19 × 10
= 10 + 190
= 200
I hope this helps you.
: )
Let a , d are first term and common
difference of an A.P
Sum of ' n ' terms = Sn
Sn = n/2 [ 2a + ( n - 1 )d ]
According to the problem given ,
S14 = 1050 , a = 10
14/2[ 2× 10 + ( 14 - 1 ) d ] = 1050
20 + 13d = 1050/7
13d = 150 - 20
13d = 130
d = 130/13
d = 10
Therefore ,
20th term in A.P = a + 19d
t20 = 10 + 19 × 10
= 10 + 190
= 200
I hope this helps you.
: )
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