Question 8
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

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Solution

$G i v e n t h a t, a_{3} = 12 a_{50} = 106 W e k n o w t h a t, a_{n} = a + (n - 1) d a_{3} = a + (3 - 1) d 12 = a + 2 d \dots \dots (i) S i m i l a r l y, a_{50} = a + (50 - 1) d 106 = a + 49 d \dots \dots (i i) O n s u b t r a c t i n g (i) f r o m (i i), w e g e t, 94 = 47 d d = 2 F r o m e q u a t i o n (i), w e g e t, 12 = a + 2 (2) a = 12 - 4 = 8 a_{29} = a + (29 - 1) d a_{29} = 8 + (28) 2 a_{29} = 8 + 56 = 64 T h e r e f o r e, 29^{t h} t e r m i s 64.$

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