Question

An A.P consists of 50 terms of which the third term is 12 and the last term is106.
Find the 29th term

Answer 1

64.. attached the picture with step by step procedure.

Answer 2

Given ,

• Number of term (n) = 50
• Third term $\tt( a_{3})$ = 12
• Last term (l) = 106

We know that , the nth term of an AP is given by

$\boxed{ \tt{ a_{n} = a + (n - 1)d}}$

Thus ,

106 = a + (50 - 1)d

106 = a + 49d --- (i)

And

12 = a + (3 - 1)d

12 = a + 2d --- (ii)

Subtract eq (ii) from eq (i) , we get

106 - 12 = a + 49d - (a + 2d)

94 = 49d - 2d

94 = 47d

d = 94/47

d = 2

Put d = 2 in eq (ii) , we get

12 = a + 2(2)

12 = a + 4

a = 8

Therefore , the first term and common difference of AP are 8 and 2

Now , the 29th term of AP will be

$\tt \implies a_{29} = 8 + (29 - 1)2$

$\tt \implies a_{29} = 8 + 28(2)$

$\tt \implies a_{29} = 8 + 56$

$\tt \implies a_{29} = 64$

Therefore , the 29th term of AP is 64

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