Question

please answer this question ​

Answer 1

Given

• Sum of first 7 terms of AP= 49
• Sum of first 17 terms of AP=289

To find :-

• We have to find the sum of n terms of AP

Solution

$\boxed{\sf\ S_n= \dfrac{n}{2}\big\{2a+(n-1)d\big\}}$

$:\implies\sf\ S_7= \dfrac{7}{2}\big\{(2a+(7-1)d \big\}\\ \\ \\ :\implies\sf\ \cancel{49}= \dfrac{\cancel{7}}{2}\big\{2a+6d\big\}\\ \\ \\ :\implies\sf\ 7= \dfrac{\cancel {2}(a+3d)}{\cancel{2}}\\ \\ \\ :\implies\sf\ a+3d=7\ ------ eq.(i)\\ \\ \\:\implies\sf\ S_{17}= \dfrac{17}{2}\big\{(2a+(17-1)d \big\}\\ \\ \\ :\implies\sf\ \cancel{289}= \dfrac{\cancel{17}}{2}\big\{2a+16d\big\}\\ \\ \\ :\implies\sf\ 17= \dfrac{\cancel {2}(a+8d)}{\cancel{2}}\\ \\ \\ :\implies\sf\ a+8d=17\ ------ eq.(ii)\\ \\ \\ \bullet\sf Subtract\ eq.(i)\ from\ (ii)\\ \\ \\ :\implies\sf\ a+8d-(a+3d)=17-7\\ \\ \\ :\implies\sf\ a+8d-a-3d=10\\ \\ \\ :\implies\sf\ 5d=10\\ \\ \\ :\implies\sf\boxed{\sf\ d=2}$

• Put value of d in equation (i)

$\bullet\sf\ a+3(2)=7\\ \\ \implies\sf\ a=7-6\\ \\ :\implies\boxed{\sf\ a=1}$

• Now the sum of n terms

$:\implies\sf\ S_n= \dfrac{n}{2}\big\{2(1)+(n-1)2\big\}\\ \\ \\ :\implies\sf\ S_n= \dfrac{n}{2}\big\{2+2n-2\big\}\\ \\ \\ :\implies\sf\ S_n= \cancel{\dfrac{2n^2}{2}}\\ \\ \\ :\implies\sf\boxed{\purple{\sf\ S_n=n^2}}$

Answer 2

The answer is in the attachment.

Please go through it.

Formulas :

• The n th term of A.P=an = a + (n – 1) × d
• Sum of n terms in A.P=S = n/2[2a + (n − 1) × d]