Question

In an AP
i)given a=7,a13=35, find d and S13.
ii)given a3=15,S10=125,find d and a10.

Answer 1

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

Step-by-step explanation:

(i)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

(ii)an=a+(n-1)d

a3=a+(3-1)d

15=a+2d

a+2d=15 _________ {1}

Sn=n/2(2a+{n-1}d)

S10=10/2(2a+{10-1}d)

125=5(2a+9d)

125/5=2a+9d

25=2a+9d ___________{2}

solving eq{1} & eq{2}

putting a value in eq {2}

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

30-4d+9d=25

5d=25-30

=-5

=d=-1

=>an=a+(n-1)d

a10=a+(10-1)d

a10=a+9d

a10=17+9(-1)

a10=17-9

a10=8