An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2013 in which we have been asked to find the AP, if the 19th term of an AP is equal to 3 times its 6th term. If it’s 9th term is 19.
RS Aggarwal, class 10, chapter 5A, question no 39
t19=3t6
⇒3(a+(6−1)d)=a+(19−1)d
3(a+5d)=a+18d
3a+15d=a+18d
2a=3d
⇒d=2/3a
⇒t9=19
a+(9−1)d=19
a+8d=19
a+8×2/3a=19
3a+16a=19×3
19a=19×3
⇒a=3
⇒d=2/3a=2/3×3=2
∴ A.P is 3,5,7,9......
Hence, solved.