Question

In an arithmetic progression 6th term is 1 more thant twice the third term the sum of the fourth and fifth term is five times the second term find the tenth term of an arithmetic progression​

Answer 1

Answer:

Step-by-step explanation:

Let a6= 2 (a3)  +1

an = a + d (n -1)

a6 = a + d (6 - 1)

a6 = a +5d

a + 5d = 2 (a3) + 1

a3 = a + 2d

∴ a + 5d = 2 (a + 2d ) +1

  a + 5d = 2a + 4d +1

 5d - 4d = 2a - a + 1

       d    = a + 1

a4+a5 =5 (a2)

(a+3d) + (a+4d) = 5 (a+d)

2a+7d = 5a+5d

7d-5d = 5a-2a

2d = 3a

Since d = a +1,

2 (a+1) = 3a

2a+2=3a

3a-2a=2

 a =2

Substitute a = 2 into 'd = a + 1'

d=2+1

Hence, d=3

a = 2, d = 3,

a10 = a + 9d

     = 2 + 9 (3)

     = 2 + 27

     = 29

• Sixth term is 1 more than twice third term:  a3= 8,  a6=17

• Sum of fourth and fifth is five times second term :

                  a4 = 11, a5= 14, 5 (a2) = 11+14=25

                 Hence, a2=5 because 5×5=25

∴ The AP is  2, 5, 8, 11, 14, 17, 20, 23, 26, 29

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Answer 2

Answer:

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