Question

In an arithmetic sequence the sum of the first n

Answer 1

Answer:

$a_{103}=195$

Step-by-step explanation:

Sum of the first n terms of an arithmetic series formula:

$\large\boxed{S_n=\dfrac12n\left[2a+(n-1)d\right]}$

where:

• a is the first term.
• d is the common difference between terms.

Given:

$S_n=n^2-10n$

Substitute the given equation into the formula and rearrange:

$\begin{aligned}\implies n^2-10n & =\dfrac12n[2a+(n-1)d]\\n(n-10)& = \dfrac{1}{2}n\left[2a+dn-d\right]\\n-10_&=\dfrac{1}{2}[2a+dn-d]\\2n-20& = 2a+dn-d\\ 2n-20 & = dn-(d-2a) \end{aligned}$

Therefore:

$\implies 2n =dn$

$\implies d = 2$

Therefore:

$\implies -20=-(d-2a)$

$\implies 20=d-2a$

Substitute the found value of d:

$\implies 20=2-2a$

$\implies 18=-2a$

$\implies 2a=-18$

$\implies a=-9$

Check the found values of a and d by substituting them into the sum formula and rearranging:

$\begin{aligned}\implies S_n & =\dfrac12n[2a+(n-1)d]\\& =\dfrac12n[2(-9)+(n-1)(2)]\\& =\dfrac12n[-18+2n-2]\\& =n[-9+n-1]\\& =-9n+n^2-n\\& = n^2-10n\end{aligned}$

As the equation matches the given equation, this verifies that:

• First term = -9
• Common difference = 2

General form of an arithmetic sequence:

$\large\boxed{a_n=a+(n-1)d}$

where:

• $a_n$ is the nth term.
• a is the first term.
• d is the common difference between terms.

To find the 103rd term, substitute the found values of a and d together with n = 103 into the formula:

$\implies a_{103}=-9+(103-1)(2)$

$\implies a_{103}=-9+(102)(2)$

$\implies a_{103}=-9+204$

$\implies a_{103}=195$

Therefore, the 103rd term of the arithmetic sequence is 195.

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