Question

ii) Find the A.P. whose 1st term is 100 and the sum of the first six terms is 5 times the sum of the next six terms.​

Answer 1

$\Hugea_{n}=100-10(n-1)$

$\huge\textbf{Steps to Solution}$

$\Large\textrm{Let us denote: -}$

$\textrm{ S_{n} , which is the sum of first n terms}$

$\textrm{ a_{n} , which is the n th term}$

$\Large\textrm{It is given that: -}$

$\text{ a_{1}=100 , S_{6}=5(S_{12}-S_{6}) }$

$\Large\textrm{It follows that: -}$

$6S_{6}=5S_{12}$

$6\times\dfrac{6(200+5d)}{2}=5\times\dfrac{12(200+11d)}{2}$

$18\times(200+5d)=30\times(200+11d)$

$3(200+5d)=5(200+11d)$

$600+15d=1000+55d$

$15d-55d=1000-600$

$-40d=400$

$d=-10$

$\Large\textrm{So: -}$

$\textrm{ \boxed{a_{n}=100-10(n-1)} is the correct answer.}$

$\huge\textbf{Let us Verify}$

$\textrm{100, 90, 80, 70, 60, 50 and 40, 30, 20, 10, 0, -10}$

$\textrm{Each sum is 450 and 90. \underline{Hence verified.}}$

$\huge\textbf{Formulae Used}$

$\Large\textrm{ \bigstar Series of A.P - (I) \bigstar }$

$\cdots\longrightarrow\boxed{S_{n}=\dfrac{n\{2a+(n-1)d\}}{2}}$

$\textrm{It is obtained by the following equation, since l=a+(n-1)d .}$

$\Large\textrm{ \bigstar Series of A.P - (II) \bigstar }$

$\cdots\longrightarrow\boxed{S_{n}=\dfrac{n(a+l)}{2}}$