Question

The 12 term of an ap exceeds the 5 term by 14 sum of both the term is 36 find 1 term an common difference

Answer 1

T12 = T5 + 14            (Given)

T12 - T5 = 14....................................................eq1

T12 + T5 = 36     (Given)....................................................eq2

Solving eq1 and eq2 we get

T12 = 25    and      T5 = 11

Now

T12 = a + (12 - 1) d                                     (Since Tn = a + (n - 1)d)

25 = a + 11d...........................................................eq3

T5 = a + (5 - 1)d

11 = a + 4d..............................................................eq4

Solving eq3 and eq4 we get

a= 3                       d =2

Happy Learning :)