Question

the sum of an 6th term of an AP is 39 and the common difference is 3 write its a6 terms​

Answer 1

Answer:

0.5, 3.5, 6.5, 9.5, 12.5 and so on...

Step-by-step explanation:

Given

n = 6

S₆ = 39

d = 3

Fina all the 6 terms of the A.P.

a₆ = a₁ + (n-1)d = a₁ + (5-1)3 = a₁ + 4(3) = a1 + 12

⇒ a₆ - a₁ = 12 - eqn.1

Now,

S₆ = (6/2)(a₆ + a₁) = 3(a₆ + a₁)

⇒ 39/3 = a₆ + a₁

⇒ a₆ + a₁ = 13 - eqn.2

Adding equations 1 and 2,

a₆ - a₁ + a₆ + a₁ = 12 + 13

⇒ 2a₆ = 25

⇒ a₆ = 25/2 = 12.5

Put the value of a₆ in eqn.2, we have

a₁ = 13-12.5 = 0.5

Now, we have

The first term 0.5, The second term 0.5 + 3 = 3.5, the third term 3.5+3 = 6.5 and so on to form the following AP.

0.5, 3.5, 6.5, 9.5, 12.5 and so on...

Answer 2

$\huge \bold \purple{heya \: friend}$

We know that

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

Now, here

Sn=39, n=6 and d=3

So,

39=6/2{2a+(6-1)3}

39=3(2a+15)

39=6a+45

39-45=6a

-6=6a

a=-1

Now, sixth term,a6:

a6=a+5d=-1+5(3)

=-1+15

=14=>Answer