Question

if the first term of an AP is 2 than solve t10-t5 is 5 time the coman difference ​

Answer 1

First term= 2

$T_{10}$ = a+(n-1)d

     = 2+(10-1)d

     = 2+9d

$T_{5}$ = a+(n-1)d

    = 2+(5-1)d

    = 2+4d

Now,

$T_{10}$$- T_{5}$

= (2+9d)-(2+4d)

= 2+9d-2-4d

= 5d

Hence proved.

Answer 2

Step by step explanation :

In an AP

a = t1 = 2

let the common difference be 'd'

t10 - t5 = 5d ...........given

Formula : tn = a + (n-1) d

therefore, t10 = 2 + (10-1) d

= 2 + 9d .....1)

Similarly, t5 = 2 + (5-1) d

= 2 + 4d ....2)

therefore,

(2 + 9d) - (2 + 4d) = 5d ...... from 1) and 2)

2 + 9d - 2 - 4d = 5d

5d = 5d

d = 1