Question

For an AP 3rd term=13, 6th term=25, its 11th term is

Answer 1

38 is answer

hope it helps you

Answer 2

Answer:

1696

Step-by-step explanation:

T3=1, T6=-11

Tn=a+(n-1)d

T3=a+2d=1——->1

T6=a+5d=-11———→2

2–1

a+5d-a-2d=-11–1

==>3d=-12

d=-4 sustitute the d value in equation 1,

a-8=1

a=9

Now we have to calculate the sum of first 32 terms

S =(n/2)[2a + (n- 1)d]

S=32/2[2*9+(31*-4)]

S=-1696.