An important question from arithmetic progression chapter as it was already asked in previous year paper of 2014 in which we have given it’s 7th term and 13th term and we have to find its AP?
RS Aggarwal, class 10, Arithmetic Progression, chapter 5A, question no 26
nth term of an AP is ,
an=a+(n−1)d
where,
a=first term
d=common difference
now, the two terms given are a7=−4 and a13=−16
so,
a+6d=−4 ……….(1)
a+12d=−16 …….(2)
Substract equation (1) from (2),
6d=−12
d=−2
From equation (1),
a+6d=−4
a+6(−2)=−4
a−12=−4
a=8
so,
a2=a+d=8−2=6
a3=a+2d=8−4=4
Therefore, the AP is,
8,6,4,2,0,.....