Question

Find the first term and common difference of an arithmetic progression are 1 and 4 respectively. Whose sum is 190 find the number of terms in that arithmetic progression

Answer 1

AnswEr :

$\bf{ Given}\begin{cases}\sf{First \: Term (a) = 1}\\\sf{Common \:Difference (d) = 4}\\ \sf{Sum \:of \:Terms(S_n) = 190}\\\sf{Find \:Number\: of \:Terms.} \end{cases}$

• Let's Head to the Question Now :

For any Arithmetic Progression ( AP ), the sum of n terms is Given by :

$\maltese\quad\large\boxed{\sf S_n = \dfrac{n}{2}\bigg(a + l\bigg)}$

where,

• n = no. of terms
• a = First Term
• l = Last Term

Since, the last term is also the nth term :

◗ l = a + (n – 1)d

• Substituting for l in the formula for sum :

$\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(a + [a + (n -1)d]\bigg)\\ \\\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(2a + (n -1)d\bigg) \\ \\\longrightarrow \tt 190 = \dfrac{n}{2}\bigg(2(1) + (n -1)4\bigg) \\ \\\longrightarrow \tt 190= \dfrac{n}{2}\bigg(2 +4n - 4\bigg) \\ \\\longrightarrow \tt190 = \dfrac{n}{2} (4n - 2) \\ \\\longrightarrow \tt190 = \dfrac{n}{\cancel2}\times \cancel2(2n - 1)\\ \\\longrightarrow \tt190 = n(2n - 1) \\ \\\longrightarrow \tt190 = 2 {n}^{2} - n \\ \\\longrightarrow \tt2 {n}^{2} - n - 190 = 0 \\ \\\longrightarrow \tt2 {n}^{2} - 20n + 19n - 190 = 0 \\ \\\longrightarrow \tt2n(n - 10) + 19(n - 10) = 0 \\ \\\longrightarrow \tt(2n + 19)(n - 10) = 0 \\ \\\longrightarrow \red{\tt n = \dfrac{ - 19}{2}} \quad \tt or \quad \green{n = 10}$

◗ we will ignore Negative Term, taking positive value of n = 10

⠀

∴ Therefore, Number of terms will be 10.

#answerwithquality #BAL

Answer 2

$\bf{\Huge{\underline{\boxed{\bf{\green{ANSWER\::}}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}}$

The first term and common difference of an arithmetic progression are 1 and 4 respectively. Whose sum is 190.

$\bf{\Large{\underline{\bf{To\:find\::}}}}}$

The number of terms in that arithmetic progression.

$\bf{\Large{\underline{\boxed{\rm{\blue{Explanation\::}}}}}}$

We know that formula of the sum of A.P.;

$\longmapsto{\bf{\sf{\orange{Sn=\frac{n}{2}[2a+(n-1)d]}}}}}}$

$\bf{We\:have\begin{cases}\rm{The\:first\:term\:(a)=1}\\ \rm{The\:common\:difference\:(d)=4}\\ \sf{The\:sum\:(Sn)=190}\end{cases}}$

Therefore,

$\leadsto\rm{Sn\:=\:\frac{n}{2} [2a+(n-1)d]}$

$\leadsto\rm{190\:=\:\frac{n}{2} [2(1)+(n-1)4]}$

$\leadsto\rm{190\:=\:\frac{n}{2} [2+4n-4]}$

$\leadsto\rm{190\:=\:\frac{n}{2} [4n-2]}$

$\leadsto\rm{380\:=\:n(4n-2)}$

$\leadsto\rm{380\:=\:4n^{2} -2n}$

$\leadsto\rm{4n^{2} -2n-380=0}$

$\leadsto\rm{2(2n^{2} -n)-380=0}$

$\leadsto\rm{2n^{2} -n-\cancel{\frac{380}{2}} =0}$

$\leadsto\rm{2n^{2} -n-190=0}$

$\bf{\Large{\boxed{\sf{Factorization\:Method\::}}}}}}$

$\leadsto\rm{2n^{2} -20n+19n-190=0}$

$\leadsto\rm{2n(n-10)+19(n-10)=0}$

$\leadsto\rm{(n-10)(2n+19)=0}$

Now,

$\longmapsto\rm{(n-10)=0\:\:\:\:\:\:\:or\:\:\:\:\:(2n+19)=0}$

$\longmapsto\rm{n\:=\:10\:\:\:\:\:\:\:or\:\:\:\:\:2n=-19}$

$\longmapsto\rm{n\:=\:10\:\:\:\:\:\:or\:\:\:\:\:n=\cancel{\frac{-19}{2} }}$

$\longmapsto\rm{n\:=\:10\:\:\:\:\:\:\:or\:\:\:\:\:\:n=-9.5}$

We know that negative value isn't acceptable, so;

$\sf{\pink{{The \:number\: of \:terms \:in \:that\: A.P. \:is \:n=10}}}$