Question

in the arithmetic progression a16=10 and S16= 200 find a11​

Answer 1

Answer:

a11=11.7 (approx.)

Step-by-step explanation:

Given:

a16=10

S16=200

To find: a11

proof:

we know that,

• an=a+(n-1)*d

      a16=a+15d

     a+15d=10---> 1

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

      S16 = 16/2(2a+15d)

      200 = 8(2a+15d)

      200/8 = 2a+15d

   2a+15d = 25----> 2

Subtract equation 1 from 2

2a+15d=25

 a+15d=10

--------------------

    a = 15

substitute a=15 in equation 1

a+15d = 10

15+15d = 10

15d = 10 - 15

15d = -5

d = -5/15

d = -1/3

we need to find 11th term,

a11 = a+10d------>3

substitute a=15 and d=-1/3 in equation 3

a11 = 15+10*(-1/3)

a11 = 15+(-10/3)

a11 = (45-10)/3

a11 = 35/3

a11 = 11.66 = 11.7 (approx.)

Therefore, a11 = 11.7

Hope it helps