Question

In an AP if sum of its first n terms is 3n2 + 5n and its k
th term is 164, find the value of k.


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Answer 1

Answer:

given Sₙ = 3n² + 5n

Step-by-step explanation:

aₙ = Sₙ - Sₙ₋₁

=> aₙ = 3n² + 5n - [ 3(n-1)² + 5(n-1)]

aₙ = 3n² + 5n - [ 3n² + 3 - 6n + 5n - 5 ]

=> 3n² + 5n - [ 3n² - n - 2 ]

=> 3n² + 5n - 3n² + n + 2

=> 6n + 2

therefore nth term aₙ = 6n + 2

given kth term ak = 164, to find k

Ak = 6k + 2

164 = 6k + 2

=> 6k = 164 - 2

=> 6k = 162

=> k = 27

Answer 2

Answer:

Hope it helps you.....

Step-by-step explanation:

Question:

In an AP if sum of its first n terms is 3n2 + 5n and its $k^{th}$ term is 164, find the value of k.

Solution:

$\bold{S_n=3n^2+5n}$

$\sf{S_1=a_1=3(1)2+5(1)=8}$

$\mathtt{S_2=3(2)2+5(2)=22}$

$\frak{s_2=22=a_1+a_2}$

$\rm{a_2=22-8=14}$

$d=a_2-a_1=14-8=6$

kth term value is 164,then what is k?

kth term=a+(k-1)d

164=8+(k-1)6

$\displaystyle\frac{(164-8)}{6}$=k-1

k-1= $\displaystyle\frac{156}{6}$

26+1=k

$\boxed{\rm{\green{k=}{ \frak{\red{27}}}}} \:$

$So\:27^{th}term\:is\:164.$