Question

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Answer 1

Answer:

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Answer 2

Answer:

The sum of first ‘n’ terms of the AP is n² when it’s sum of first 7 terms is 49 and that of 17 terms is 289.

$\qquad$_____________

Step-by-step explanation:

According to the question, it is given that the sum of first 7 terms of the AP is 49. So :

$\tt\implies~ S_n=\dfrac{n}{2}~~[2a+(n-1)d]$

$\tt\implies~ S_7=\dfrac{7}{2}~~[2a+(7-1)d]$

$\tt\implies~ S_7=\dfrac{7}{2}~~[2a+6d]$

$\tt\implies~ 49=\dfrac{7}{2}~~[2a+6d]$

$\tt\implies~ 49\times\dfrac{2}{7}=2a+6d$

$\tt\implies~ \cancel{49}\times\dfrac{2}{\cancel 7}=2a+6d$

$\tt\implies~ 7\times 2=2a+6d$

$\tt\implies~ 14=2a+6d$

$\tt\implies~ 14=2~~(a+3d)$

$\tt\implies~ \dfrac{14}{2}=a+3d$

$\tt\implies~ \cancel\dfrac{14}{2}=a+3d$

$\tt\implies~ 7=a+3d~~~\dots\dots(eq^n.I)$

$\:$

Also, it is given that, the sum of 17 terms is 289. So, we get :

$\tt\implies~ S_n=\dfrac{n}{2}~~[2a+(n-1)d]$

$\tt\implies~ S_{17}=\dfrac{17}{2}~~[2a+(17-1)d]$

$\tt\implies~ S_{17}=\dfrac{17}{2}~~[2a+16d]$

$\tt\implies~ 289=\dfrac{17}{2}~~[2a+16d]$

$\tt\implies~ 289\times\dfrac{2}{17}=2a+16d$

$\tt\implies~ \cancel{289}\times\dfrac{2}{\cancel{17}}=2a+16d$

$\tt\implies~ 17\times 2=2a+16d$

$\tt\implies~ 34=2a+16d$

$\tt\implies~ 34=2~~(a+8d)$

$\tt\implies~ \dfrac{34}{2}=a+8d$

$\tt\implies~ \cancel\dfrac{34}{2}=a+8d$

$\tt\implies~ 17=a+8d~~~\dots\dots(eq^n.II)$

$\:$

Now, let’s subtract eq . I from eq . II to get the common difference ‘d’ :

$\tt\implies~ 17-7=a+8d-(a+3d)$

$\tt\implies~ 10=a-a+8d-3d$

$\tt\implies~ 10=5d$

$\tt\implies~ \dfrac{10}{5}=d$

$\tt\implies~ \cancel\dfrac{10}{5}=d$

$\tt\implies~ d=2$

$\:$

Substituting this value of d in eq . I to get the value of a :

$\tt\implies~ 7=a+3d$

$\tt\implies~ 7=a+3~(2)$

$\tt\implies~ 7=a+6$

$\tt\implies~ 7-6=a$

$\tt\implies~ a=1$

$\:$

Finally, let’s substitute these values of ‘a’ and ‘d’ to get the sum of ‘n’ terms :

$\tt\implies~ S_n=\dfrac{n}{2}~~[2a+(n-1)d]$

$\tt\implies~ S_n=\dfrac{n}{2}~~[2(1)+(n-1)(2)]$

$\tt\implies~ S_n=\dfrac{n}{2}~~[2+2n-2]$

$\tt\implies~ S_n=\dfrac{n}{2}~~[2n]$

$\tt\implies~ S_n=\dfrac{n}{\cancel2}~~[\cancel{2n}]$

$\tt\implies~ S_n=n\times n$

$\tt\implies~ S_n=n^2$

$\:$

$\qquad$_____________

Therefore, the sum of ‘n’ terms of AP is n².