Question

Sum of first ‘n' natural numbers is 231. What number is ‘n' term ?

Answer 1

Answer:

$\sf{The \ number \ of \ terms \ is \ 21.}$

Given:

• The sum of 'n' terms is 231.

To find:

• The number of 'n' terms.

Solution:

$\sf{Here,}$

$\sf{\longmapsto{First \ term \ (a)=1}}$

$\sf{...[\because{Natural \ number \ starts \ from \ 1}]}$

$\sf{\longmapsto{Common \ difference \ (d)=2-1=1}}$

$\sf{\longmapsto{Sum \ of \ n \ terms \ (S_{n})=231}}$

$\sf{It \ forms \ an \ A.P.,}$

$\boxed{\sf{S_{n}=\dfrac{n}{2}[2a+(n-1)d]}}$

$\sf{\therefore{231=\dfrac{n}{2}[2(1)+(n-1)\times1]}}$

$\sf{\therefore{231\times2=n(2+n-1)}}$

$\sf{\therefore{462=n(n+1)}}$

$\sf{\therefore{n^{2}+n-462=0}}$

$\sf{Here, \ a=1, \ b=1 \ and \ c=-462}$

$\sf{\mapsto{b^{2}-4ac=1^{2}-4(1)(-462)}}$

$\sf{\mapsto{b^{2}-4ac=1+1848}}$

$\sf{\mapsto{b^{2}-4ac=1849}}$

$\sf{By \ quadratic \ formula}$

$\sf{\leadsto{n=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}$

$\sf{\leadsto{n=\dfrac{-1\pm\sqrt{1849}}{2(1)}}}$

$\sf{\leadsto{n=\dfrac{-1\pm43}{2}}}$

$\sf{\leadsto{n=\dfrac{-1+43}{2} \ or \ \dfrac{-1-43}{2}}}$

$\sf{\leadsto{n=\dfrac{42}{2} \ or \ \dfrac{-44}{2}}}$

$\sf{\leadsto{n=21 \ or \ -22}}$

$\sf{But, \ number \ of \ terms \ can't \ be \ negative.}$

$\sf{\therefore{n\neq \ -22}}$

$\boxed{\sf{\therefore{n=21}}}$

$\sf\purple{\tt{\therefore{The \ number \ of \ terms \ is \ 21.}}}$