Question

7, 9, 13 are the first three terms of a sequence with

Answer 1

(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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