Bq)(ii) 27a313 - 18a-b3 + 7503b2(ii) 39(x2 + y2) + 6b(x2 + y2)35/2

1.	Bq)(ii) 27a313 - 18a-b3 + 7503b2(ii) 39(x2 + y2) + 6b(x2 + y2)35/2
Answer» Answer: LET p, be the first term l.e; 18 And, d, be the common difference, From given series 4th term = p+3D -3 18+ 3d = -3 Hence, d= -7 So, a = p+d = 18 – 7 = 11 And, b=p+ 2d=18 - 14 = 4 Therefore, a+b= 11+4=15

Answer»

Answer:

LET p, be the first term l.e; 18

And, d, be the common difference,

From given series

4th term = p+3D -3

18+ 3d = -3

Hence, d= -7

So, a = p+d = 18 – 7 = 11

And, b=p+ 2d=18 - 14 = 4

Therefore,

a+b= 11+4=15

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Bq)(ii) 27a313 - 18a-b3 + 7503b2(ii) 39(x2 + y2) + 6b(x2 + y2)35/2​

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Bq)(ii) 27a313 - 18a-b3 + 7503b2(ii) 39(x2 + y2) + 6b(x2 + y2)35/2