Question

The first term and the common difference of an AP are 1 and 4 respectively .whose sum is 190 .find the number of terms in that AP

Answer 1

so there are 10 terms in the given AP

Hope it's helpful......

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;

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