Question

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

Answer 1

Answer:

the 29th term is 64

Step-by-step explanation:

We know that the formula for the nth term is t

n

=a+(n−1)d, where a is the first term, d is the common difference.

It is given that third term of an A.P is t

3

=12, therefore,

t

n

=a+(n−1)d

⇒12=a+(3−1)d

⇒a+2d=12......(1)

Also, it is given that the 50th term is t

50

=106, therefore,

t

n

=a+(n−1)d

⇒106=a+(50−1)d

⇒a+49d=106......(2)

Now, subtract equation 1 from 2 as follows:

(a−a)+(49d−2d)=106−12

⇒47d=94

⇒d=

47

94

=2

Substitute the value of the difference d=2 in equation 1:

a+(2×2)=12

⇒a+4=12

⇒a=12−4=8

Now, the 29th term with a=8 and d=2 can be obtained as:

t

29

=8+(29−1)2=8+(28×2)=8+56=64

Hence, the 29th term is 64

Answer 2

Step-by-step explanation:

$a_{3} = 12 \\a + 2d = 12 \: \: \: \: \: \: \: \: \: \: \: eq(1) \\ a_{50} = 106 \\ a + 49d = 106 \: \: \: \: \: \: eq(2) \\ eq(2) \: - \: eq(1) \\ \: a + 49d - a - 2d \: = \: 106 - 12 \\ 47d = 94 \\ d = \frac{94}{47} \\ d = 2 \\ put \: the \: value \: of \: d \: in \: eq \: (1) \\ a + 2 \times 2 = 12 \\ a + 4 = 12 \\ a = 12 - 4 \\ a = 8\\ now \\ a_{29} = a + 28d = 8 + 28 \times 2 \\ = 8 + 56 = 64 \\ a_{29} = 64$

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