Question

In an A.P the sum of first n terms is 3n^2+5n and its kth term is 164. Find the value of k​

Answer 1

Answer:

k = 27

Step-by-step explanation:

Given, Sum of first n terms is 3n² + 5n.

Sn = 3n² + 5n

Substitute n = 1, we get

T₁ = S₁ = 3 + 5 = 8

Substitute n = 2, we get

S₂ = T₁ + T₂ = 12 + 10 = 22

Then, T₂ = S₂ - T₁

               = S₂ - S₁

               = 22 - 8

               = 14

Common difference d = T₂ - T₁

                                     = 14 - 8

                                     = 6

Here, First term a = 8, Common difference d = 6.

Now,

Given kth term is 164.

⇒ a + (k - 1) * d = 164

⇒ 8 + (k - 1) * 6 = 164

⇒ 8 + 6k - 6 = 164

⇒ 6k + 2 = 164

⇒ 6k = 162

⇒ k = 27.

Therefore, the value of k = 27.

Hope it helps!

Answer 2

Given: Sn = 3n² + 5n

S₁ = 3(1)² + 5(1)

= 3 + 5

T₁ = 8

S₂ = 3(2)² + 5(2)

= 12 + 10

= 22

So, T₁ + T₂ = 22

8 + T₂ = 22

T₂ = 22 - 8

T₂ = 14

Now,

a = 8 and d = 14 - 8 = 6

As ak = 164

or a + (k-1)d = 164

8 + (k-1)6 = 164

6k - 6 = 164 - 8

6k = 164 + 6 - 8

6k = 162

k = 162/6

k = 27