an A.P consists of 50terms of which 3rd term is 12 and last digit is 106. find its 29th term
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Given that,
a3 = 12
a50 = 106
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d (I)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d (II)
On subtracting (I) from (II), we obtain
94 = 47d
d = 2
From equation (I), we obtain
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29th term is 64.
@skb
a3 = 12
a50 = 106
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
12 = a + 2d (I)
Similarly, a50 = a + (50 − 1) d
106 = a + 49d (II)
On subtracting (I) from (II), we obtain
94 = 47d
d = 2
From equation (I), we obtain
12 = a + 2 (2)
a = 12 − 4 = 8
a29 = a + (29 − 1) d
a29 = 8 + (28)2
a29 = 8 + 56 = 64
Therefore, 29th term is 64.
@skb
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