Question

An AP consists of 50 terms of which 3rd term is 12 and last term is 106. Find sum of its 29 terms.

Answer 1

Answer:  

134450/47

Step-by-step explanation:

$a_{3} = 12\\a_{1} + 2d = 12    -(1)\\a_{50} = 106\\a_{1} + 49d = 106 -(2)\\\\\\subtract (1) and (2)\\\\a_{1} + 2d = 12   \\a_{1} + 49d = 106 \\--------\\47d = 84\\d = 84/47     \\  \\put in (2)\\a_{1} = 12-2*84/47\\a_{1} = 12- 168/47\\a_{1} = 564- 168/47$$a_{1} = 396/47$$Sn = n/2 (a_{1} + a_{50} )\\         50/2 ( 396/47+106 )\\      25( 396/47 + 4982/47 )\\     25(5378/47)\\134450/47$

Answer 2

Answer:

Step-by-step explanation:

Hi ,

Let a , d are first term and common

difference of an A.P

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nth term = Last term = a + ( n - 1 )d

an = a + ( n - 1 )d

********************************************

Now ,

It is given that ,

Third term = 12

a + 2d = 12 ------( 1 )

Last term = 106

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

Subtract ( 1 ) from ( 2 ) , we get

47d = 94

d = 2

Substitute d value in equation ( 1 ) ,

We get

a + 2 × 2 = 12

a = 12 - 4

a = 8

29th term = a + 28d

a29 = 8 + 28 × 2

= 8 + 56

= 64

I hope this helps you.