if 9th term of an ap is zero prove that its 29th term is double of its 19th term
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Hey !!!
9th term of an AP is 0
tn = 0
n = 9
As we know that
tn = a+ (n-1)d
t9 = a + (9-1)d
0 = a + 8d
a = -8d -------1)
now , proofe :- 2t19 = t29
t19 = a + 18d
t19 = -8d + 18d
t19 = 10d [from 1 ]
and
t29 = a + 28d
t29 = -8d + 28d
t29 = 20d
now, if we multiply by 2 in t19 then here prooved
2t19 = t29
2× 10d = 20d
20d = 20d
LHS = RHS so prooved here
2t19 = t29
Hope it helps you !!!
@Rajukumar111
9th term of an AP is 0
tn = 0
n = 9
As we know that
tn = a+ (n-1)d
t9 = a + (9-1)d
0 = a + 8d
a = -8d -------1)
now , proofe :- 2t19 = t29
t19 = a + 18d
t19 = -8d + 18d
t19 = 10d [from 1 ]
and
t29 = a + 28d
t29 = -8d + 28d
t29 = 20d
now, if we multiply by 2 in t19 then here prooved
2t19 = t29
2× 10d = 20d
20d = 20d
LHS = RHS so prooved here
2t19 = t29
Hope it helps you !!!
@Rajukumar111
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