Question

50 points..An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. No unnessasary answering. Be like a person and don't simply verge around...​

Answer 1

Answer:

                                                                                 

Given, a3 = 12 and a50 = 106

a3 = a + 2d = 12

a50 = a + 49d = 106

Subtracting 3rd term from 50th term, we get;

a + 49d – a – 2d = 106 – 12

Or, 47d = 94

Or, d = 2

Substituting the value of d in 12th term, we get;

a + 2 x 2 = 12

Or, a + 4 = 12

Or, a = 8

Now, 29th term can be calculated as follows:

a29 = a + 28d

= 8 + 28 x 2

= 8 + 56 = 64

hope it helps you

                                                                                     

Answer 2

$\underline{\underline{\textbf{Answer}}}$

$\huge{\boxed{\boxed{\red{64}}}}$

Number of terms in the A.P = 50 .

Let the 1 st term be a .

Let the common difference be d .

n th term is given by a + ( n - 1 ) d

3 rd term will hence be a + ( 3 - 1 ) d

3 rd term = a + 2 d

Given the 3 rd term is 12 .

So a + 2 d = 12 ............(1)

Given the last term is 106 .

So the 50 th term is 106

a + ( 50 - 1 ) d = 106

a + 49 d = 106 ...............(2)

We got 2 equations and we will solve them by subtracting the equation (2) from the equation (1)

2 d - 49 d = 12 - 106

⇒ - 47 d = - 94

⇒ d = $\frac{-94}{-47}$

⇒ d = 2

So the common difference will be 2 .

Put d = 2 in equation (1)

a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 12 - 4

⇒ a = 8

The first term is 8 .

29 th term = a + ( 29 - 1 ) d

⇒ a + 28 d

⇒ 8 + 28 × 2

⇒ 8 + 56

⇒ 64

The 29 th term will be 64 .