the sixth term of an ap is five times the first one and the eleventh term exceeds twice the fifth term by 3 find its eight th term meritnation

Dear Student,

Please find below the solution to the asked query:

We know n^th term in A.P. :

T_n =  a + ( n - 1 ) d  , Here a =  first term and d =  common difference .

Given : the sixth term of an A.P. is five times the first one , So

T₆  = 5 ( T₁ )

a + ( 6 - 1 ) d = 5 [ a + ( 1 - 1 ) d ]

a + 5d =  5a

4 a  - 5d = 0                                                                              ---- ( 1 )

Also given : The eleventh term exceeds twice the fifth term by 3 , Sp

T₁₁ =  2 ( T₅ ) + 3

a + ( 11 - 1 ) d = 2 [ a + ( 5 - 1 ) d ] + 3

a + 10 d = 2 [ a + 4 d ] + 3

a + 10 d = 2 a + 8 d  + 3

- a + 2 d = 3                                                                              ---- ( 2 )

Now we multiply by 4 in equation 2 and get

- 4 a + 8 d = 12                                                                         ---- ( 3 )

Now we add equation 1 and 3 and get

3 d  = 12

d  =  4 , Substitute that value in equation 2 and get

- a + 2 ( 4 ) = 3

- a + 8 = 3

- a = - 5

a = 5

So,

8^th term :

T₈ = 5 + ( 8 - 1 ) 4 =  5 + ( 7 ) 4 =  5 + 28 =  33                                     


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