Question

If the sum of the first n terms of A.P. 30,27,24,21,...... is 120, find the number of terms and the last term.​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{A.P is 30,27,24,21,.........and sum of n terms is 12}$

$\underline{\textsf{To find:}}$

$\textsf{The number of terms and the last term}$

$\underline{\textsf{Solution:}}$

$\textsf{Concept used:}$

$\boxed{\begin{minipage}{7cm} \textsf{The sum of n terms of the A.P a, a+d,.......is}\\\\\mathsf{S_n=\dfrac{n}{2}[2a+(n-1)d]} \end{minipage}}$

$\textsf{For the given A.P a=30, d=-3 and}$

$\mathsf{S_n=120}$

$\mathsf{\dfrac{n}{2}[2a+(n-1)d]=120}$

$\mathsf{\dfrac{n}{2}[2(30)+(n-1)(-3)]=120}$

$\mathsf{\dfrac{n}{2}[60-3n+3]=120}$

$\mathsf{\dfrac{n}{2}[63-3n]=120}$

$\mathsf{\dfrac{3n}{2}[21-n]=120}$

$\mathsf{\dfrac{n}{2}[21-n]=40}$

$\mathsf{21n-n^2=80}$

$\mathsf{21n-n^2-80=0}$

$\mathsf{n^2-21n+80=0}$

$\mathsf{(n-5)(n-16)=0}$

$\mathsf{n=5,16}$

$\implies\boxed{\textsf{The number of terms is 16}}$

$\textsf{The last term}\;\mathsf{=t_{16}}$

$\mathsf{=a+15d}$

$\mathsf{=30+15(-3)}$

$\mathsf{=30-45}$

$\mathsf{=-15}$

$\implies\boxed{\textsf{The last term is -15}}$

$\textbf{Find more:}$

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