If the sum of the first $n$ terms of an A.P. is $4n – n^2$, what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the $n$th terms.

Given:

The sum of first $n$ terms of an A.P. is $4n-n^{2}$.

To do:

We have to find the first term, sum of first two terms, second term, third term, tenth term and the $n$th term. 

Solution:

$S_{n} =4n-n^{2}$

For $n=1,\ S_{1} =4\times 1 -1^2=4-1=3$

Therefore, first term $a=3$

For $n=2,\ S_{2} =4\times 2-2^{2}=8-4=4$

$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$

$=4-3$

$=1$

Common difference of the A.P., $d=$second term $-$ first term

$=1-3=-2$

We know that,

$a_{n}=a+(n-1)d$

$a_{2}=a+d=3+(-2)=3-2=1$

$a_3=a+2d=3+2(-2)=3-4=-1$

$a_{10}=a+(10-1)d=3+9(-2)=3-18=-15$

$a_n=a+(n-1)d=3+(n-1)(-2)$

$=3-2n+2$

$=5-2n$

Therefore, the first term is $3$, the second term is $1$, the third term is $-1$, the tenth term is $-15$, $n$th term is $5-2n$ and the sum of the first two terms is $4$.

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Updated on: 10-Oct-2022

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