Question

1) determine k so that k+2 , 4k-6 and 3k-2 are the three consecutive terms of an AP.

2) if 7 times the 7th term of an AP is equal to 11 times the 11th term , show that the 18th term is zero.

Answer 1

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REFER TO THE ATTACHMENT...

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Answer 2

1.Let
a-d=k+2. (1)
a=4k-6. (2)
a+d=3k-2. (3)
put(2) in (1)
4k-6-d=k+2
d=3k-8. (4)
put (4) in (3)
4k-6+3k-8=3k-2
7k-14=3k-2
4k=12
k=3

2. 7(a+6d)=11(a+10d)
7a+42d=11a+110d
-4a=68d
a=-17d
a of 18th term=a+17d
=-17d+17d
=0