Question

In an A,if a=7,a13=35find `D`ands13

Answer 1

Answer:

Step-by-step explanation:

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

  

Answer 2

Answer:

d = 7/3, Sn=273

Step-by-step explanation:

First term of an AP = a = 7

Thirteenth term of an AP = 35

a + 12d = 35 ------(1)

Substitute a in eq - (1)

a + 12d = 35

(7) + 12d = 35

12d = 35 - 7

12d = 28

d = 28/12

d = 7/3

In an AP sum of the terms = n/2 ( a + an )

= 13/2 ( 7 + 35)

= 13/2 ( 42)

= 13(21)

= 273