Answer:
Step-by-step explanation:
[tex]S_{n}=\dfrac{n}{2}(2a + (n-1)*d)[/tex]
Here, n- number of terms ; d - common difference ; a - first term
[tex]S_{8}=12\\\\\dfrac{8}{2}[2a+(8-1)*d]=12\\\\4[2a+7d]=12\\\\2a + 7d = \dfrac{12}{4}\\\\\\2a + 7d = 3 \ -----------------------(i)[/tex]
[tex]S_{16}=56\\\\\dfrac{16}{2}(2a+15d)=56\\\\\\8(2a+15d)=56\\\\2a + 15d=\dfrac{56}{8}\\\\\\[/tex]
2a + 15d = 7 ----------------(ii)
Subtract (i) from equation (ii)
(ii) 2a + 15d = 7
(i) 2a + 7d = 3
- - -
8d = 4
d = 4/8
d = 1/2
Plugin d = 1/2 in equation (i)
[tex]2a +7*\dfrac{1}{2}=3\\\\\\2a = 3 -\dfrac{7}{2}\\\\\\2a =\dfrac{6}{2}-\dfrac{7}{2}\\\\\\2a=\dfrac{-1}{2}\\\\\\a=\dfrac{-1}{2*2}\\\\\\a=\dfrac{-1}{4}[/tex]
Second term = first term + d
[tex]=\dfrac{-1}{4}+\dfrac{1}{2}=\dfrac{-1}{4}+\dfrac{2}{4}=\dfrac{1}{4}[/tex]
[tex]Third \ term =\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1}{4}+\dfrac{2}{4}=\dfrac{3}{4}\\\\\\Fourth \ term =\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{2}{4}=\dfrac{5}{4}\\\\[/tex]
AP is :
[tex]\dfrac{-1}{4}; \dfrac{1}{4};\dfrac{3}{4};\dfrac{5}{4}.......[/tex]
