Question

In an A.P if sum of its first n terms is 3n square +5n and it's Kth term is 164, find the value of k.

Answer 1

sum of first n terms

$= ( \frac{n}{2} )(a + a + (n - 1)d) \\ = ( \frac{n}{2} )(2a + (n - 1)d) \\ = an \: + \frac{ {n}^{2}d }{2} - \frac{nd}{2}$

where a = 1st term

d = common difference

sum=

$3 {n}^{2} + 5n$

$an \: + \frac{ {n}^{2}d }{2} - \frac{nd}{2}$ =$3 {n}^{2} + 5n$

equating
$\frac{ {n}^{2}d }{2} =3 {n}^{2}$
d = 6
using this value and equating rest
n(a- d/2) = 5n
a - 6/2 = 5
a = 5+3
a =8

kth term = 164

a + (k-1) d = 164
8 + (k-1)6 = 164
6(k-1) = 156
k-1 = 26
k = 27

Answer 2

Answer:

• Value of K = 27.

Step-by-step explanation:

• A.P's sum of 'n' terms = $\bold{3n^{2}+5n}$ {Given} -(i)

But we know that, Sum of 'n' terms of an A.P is $\bold{S_{n}\:=\:\frac{n}{2}(2a+(n-1)d)}$ -(ii)

∴ $\bold{\underbrace{3n^{2}+5n}_{From\:eq^{n}\:(i)}\:=\:\underbrace{\frac{n}{2}(2a+(n-1)d)}_{From\:eq^{n}\:(ii)}}$

According to General Formula given,i.e.  A.P's sum of 'n' terms = $\bold{3n^{2}+5n}$

Putting the value of n as 1 , We get :

$\bold{3n^{2}+5n}$ {n=1}

$\bold{\Longrightarrow\:3\times1^{2}+5\times1}$

$\bold{\Longrightarrow\:3\times1+5\times1}$

$\bold{Longrightarrow\:3+5}$

• $\bold{\hookrightarrow\:8\:\:\:(a_{1}\:=8)}$

Putting the value of n as 2 , We get :

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

$\bold{\Longrightarrow\:3\times2^{2}+5\times2}$

$\bold{\Longrightarrow\:3\times4+5\times2}$

$\bold{\Longrightarrow\:12+10}$

$\bold{\Longrightarrow\:22}$

Here, ∵ This is the general formula of Summation of A.P is used, Hence

$\bold{S_{2}\:=\:22}$

$\bold{\implies\:a_{1}+a_{2}\:=\:22}$

$\bold{\implies\:8+a_{2}\:=\:22}$

$\bold{\implies\:a_{2}\:=\:22-8}$

• $\bold{\hookrightarrow\:(a_{2}\:=14)}$

Putting the value of n as 3 , We get :

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

$\bold{\Longrightarrow\:3\times3^{2}+5\times3}$

$\bold{\Longrightarrow\:3\times9+5\times3}$

$\bold{\Longrightarrow\:27+15}$

$\bold{\Longrightarrow\:42}$

Here, ∵ This is the general formula of Summation of A.P is used, Hence

$\bold{S_{3}\:=\:42}$

$\bold{\implies\:a_{1}+a_{2}+a_{3}\:=\:42}$

$\bold{\implies\:22+a_{3}\:=\:42}$

$\bold{\implies\:a_{3}\:=\:42-22}$

• $\bold{\hookrightarrow\:(a_{3}\:=20)}$

∴A.P so far formed = 8, 14, 20... with Common Difference = 6

______________________________

Coming to second part of question,

Q. Kth term is 164, find the value of k.

We know the formula,

$\bold{\implies\:a_{n}\:=\:a+(n-1)d}$

$\bold{\implies\:K_{th}\:=\:a+(k-1)d}$ {Changing 'n' into 'K'}

$\bold{\implies\:K_{th}\:=\:a+(k-1)d}$

∵Kth term is 164 (Given)

$\bold{\implies\:164\:=\:a+(k-1)d}$

Values of 'a' and 'd' are 8 and 6 respectively. (Found out above)

$\bold{\implies\:164\:=\:8+(k-1)6}$

$\bold{\implies\:164-8\:=\:(k-1)6}$ {Transposing 8 to LHS}

$\bold{\implies\:156\:=\:(k-1)6}$

$\bold{\implies\:\frac{156}{6}\:=\:(k-1)}$ {Transposing 6 to LHS}

$\bold{\implies\:26\:=\:(k-1)}$

$\bold{\implies\:26+1\:=\:(k)}$ {Transposing 1 to LHS}

$\bold{\implies\:27\:=\:k}$

∴ $\bold{\boxed{Value\:of\:K\:=\:27}}$

Thanks!