I f a , b a n d c a r e i n A P t h e n a + 1 b , b + 1 c a , c + 1 a b a r e i n ( a ) A P ( b ) G P ( c ) H P ( d ) N o n e o f t h e s e Share with your friends Share 1 Varun Rawat answered this Since, a, b and c are in AP, theb - a = c - a .....1Let t1 = a + 1bc; t2 = b + 1ca; t3 = c + 1abNow, t2 - t1 = b + 1ca -a - 1bc=b-a + 1ca - 1bc=b-a + b - aabc=b-a1 + 1abcSo, t2 - t1 = b-a1 + 1abc ......2Now, t3 - t2 = c + 1ab - b - 1ca=c-b + 1ab - 1ca=c-b + c - babc=c-b1 + 1abcSo, t3 - t2 = c-b1 + 1abc = b-a1 + 1abc Using 1 ....3So, from 2 and 3, we get t2 - t1 = t3 - t2⇒t1, t2 and t3 are in AP. 0 View Full Answer