Question

in an arithmetic Progression if sums of its first n terms is 3n²+5n and its kth term is 164,
fund the value of k​

Answer 1

Answer:

27

Step-by-step explanation:

Sum of n terms = 3n² + 5n

So, sum of (n - 1) terms = 3(n - 1)² + 5(n - 1)

nth term

= Sum of n terms - sum of (n - 1) terms

= 3n² + 5n - [3(n - 1)² + 5(n - 1)]

= 3n² + 5n - [3n² + 3 - 6n + 5n - 5]

= 3n² + 5n - 3n² - 3 + 6n - 5n + 5

= 6n + 2

Therefore, kth term = 6k + 2. As given,

kth term = 164

6k + 2 = 164

6k = 162

k = 27

Answer 2

Given :-

In  an arithmetic Progression if sums of its first n terms is 3n²+5n and its kth term is 164,

To Find :-

Value of k

Solution :-

First we will find the first term of the AP

Puttting S = 1

S₁ = 3(1)² + 5(1)

S₁ = 3 × 1 + 5

S₁ =  3 + 5

S₁ = 8

S₁ = a₁ = 8

Now

Putting S = 2

S₂ = 3(2)² + 5(2)

S₂ = 3 × 4 + 10

S₂ = 12 + 10

S₂ = 22

a₁ + a₂ = 22

a₂ = 22 - a₁

a₂ = 22 - 8

a₂ = 14

Finding common difference

D = a₂ - a₁

D = 14 - 8

D = 6

Now

aₙ = a + (n - 1)d

we have n as k

164 = a + (k - 1)d

164 = 8 + (k - 1)6

164 = 8 + 6k - 6

164 = 2 + 6k

164 - 2 = 6k

162 = 6k

162/6 = k

27 = k

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