Question

write an AP whose Third term is 5 and ten term is 9​

Answer 1

Solution:-

Given:-

=> T₃ = 5

=> T₁₀ = 9

Formula:-

=> Tₙ = a + ( n - 1 )d

Now Take T₃ = 5

=> a + ( 3 - 1 )d = 5

=> a + 2d = 5 ........(i) eq

Now Take T₁₀ = 9

=> a + ( 10 - 1 )d = 9

=> a + 9d = 9 .......(ii)eq

Now subtract ( ii )eq from ( i )eq

=> a + 2d - a - 9d = 5 - 9

=> - 7d = - 4

=> d = 4/7

Now put the value of d on ( i )st equation

=> a + 2d = 5

=> a + 2 × (4/7) = 5

=> a + 8/7 = 5

=> a = 5 - (8/7)

=> a =( 5×7 - 8 )/7

=> a = (35 - 8)/7

=> a = 27/7

So first term ( a ) 27 / 7 and common difference (d) = 4/7

Now sequences are

=> 27/7 , 31/7 , 5 ,...............