Answer: put (n=1) for sum of first ,n=2 for second ,n=3 for third and so on. S1 =4(1)+3(1) =7 S2=4(2)+3(2) =14 you can find d by only two but for making conform that it is an AP so I am also taking 3rd S3=4(3)+3(3) =21 now ,a is sum of first much and A2 =S2-S1 =7 A3= S3-S2 = 7 you can se that common difference is zero d= a2-a1 =7-7 =0 hence, AP 7,7,7,7,7 Sorry, I don’t explain properly, please don’t mind Reply
Answer:
put (n=1) for sum of first ,n=2 for second ,n=3 for third and so on.
S1 =4(1)+3(1)
=7
S2=4(2)+3(2)
=14
you can find d by only two but for making conform that it is an AP so I am also taking 3rd
S3=4(3)+3(3)
=21
now ,a is sum of first much
and A2 =S2-S1
=7
A3= S3-S2
= 7
you can se that common difference is zero
d= a2-a1
=7-7
=0
hence, AP
7,7,7,7,7
Sorry, I don’t explain properly, please don’t mind