i) *given a=7, a13=35*

*a13 (or) a+12d = 35 ------- 1*

*a = 7 (or) a+0d = 7 ------- 2*

*by subtracting 1 with 2 we get*

* *

* a+12d = 35*

* a+ 0d = 07*

* - - -*

* ------------------*

* 12d = 28*

* -------------------*

* d = 28/12 = 2.33*

*so, d = 2 (approximately)*

*S13 = n/2 [a+l]*

*n=13, a= 7, a13 = l = 35*

*S13 = 13/2 [7+35]*

*S13 = 13/2 [42]*

*S13 = 13 [21]*

*therefore, S13 = 273*

*ii) S10 = 125, a3 = 15 is given*

* S10 = 125 = 10/2 (2a+9d) [since a+l means a+a10 = a+a+9d]*

* 2a+9d=25 ------- 1*

* a3 = a+2d = 15 ------ 2*

* subtracting 2 from 1 we get, *

* 2a+9d-(a+2d) = 25-15*

* a+7d=10 *

* i.e., a8 = 10 and given a3=15*

*subtracting a3 from a8 we get*

* (a+7d)-(a+2d) = 10-15*

* 5d=-5*

* d = -1*

*by keeping it in 2 we get*

* a+2(-1) = 15*

* a= 15+2*

* a= 17*

* then,*

* a10 = a+9d = 17+9(-1) = 17-9 = 8*

* therefore, *

* a10 = 8*

* *