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Question

Question 2 (iv)
Verify that each of the following is an AP and then write its next three terms.
a + b, (a + 1) + b, (a + 1) + (b + 1) . . . .

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Solution

Here, a1=a+b,a2=(a+1)+b,a3=(a+1)+(b+1)
a2a1=(a+1)+b(a+b)=a+1+ba=1
a3a2=(a+1)+(b+1)[(a+1)+b]
=a+1+b+1a1b=1
a2a1=a3a2=1=Common difference
Since the difference between successive terms are same, it forms an AP.
The next three terms are,
a4=a1+3d=a+b+3(1)=(a+2)+(b+1)
a5=a1+4d=a+b+4(1)=(a+2)+(b+2)
a6=a1+5d=a+b+5(1)=(a+3)+(b+2)


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