(i) The simultaneous equations in a, b and c are a+b+c = 7 , 4a+2b+c = 9 & 9a+3b+c = 13
(ii) Expression for nth term = n²-n+7
The first three terms of the sequence are 7, 9, 13
nth term = an²+bn+c
If we put n = 1 then,
1st term = a(1)²+b+c
⇒ a+b+c = 7 ......(1)
If we put n = 2 then,
2nd term = a(2)²+b×2+c
⇒ 4a+2b+c = 9 ........(2)
3rd term = a(3)²+b×3+c
⇒ 9a+3b+c = 13 .......(3)
Multiplying 4 with (1) and subtract (2) from (1) we get,
(4a+4b+4c)-(4a+2b+c)=28-9
⇒ 4a+4b+4c-4a-2b-c=19
⇒ 2b+3c=19 .....(4)
Multiplying 9 with (1) and subtract (3) from (1) we get,
(9a+9b+9c)-(9a+3b+c)=63-13
⇒ 9a+9b+9c-9a-3b-c=50
⇒ 6b+8c=50
⇒ 3b+4c=25 .....(5)
Multiplying 3 with (4) and multiplying 2 with (5), then subtract (5) from (4),
3(2b+3c)-2(3b+4c)=57-50
⇒ 6b+9c-6b-8c=7
⇒ c = 7
From (5),
3b+28 = 25
⇒ 3b = -3
⇒ b = -1
From (1),
a+(-1)+7 = 7
⇒ a-1=0
⇒ a=1
So, nth term = n²-n+7
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