Question

an ap consists of 21 terms. sum of 3terms in middle is 129 and of the last 3terms is 237.find the ap

Answer 1

n=21
Sum of 3 middle terms=129
Sum of last 3 terms=237

3 middle terms of the AP=$\frac{21-3}{2}+1, \frac{21-3}{2} +2, \frac{21-3}{2} +3$=10$^{th}$, 11$^{th}$, 12$^{th}$

Sum of 3 middle terms of the AP=$a_{10} + a_{11}+ a_{12} =a+9d+a+10d+a+11d$
3a+30d=129
a+10d=43
a=43-10d  ...(1)

Sum of 3 terms of the AP=$a_{19}+ a_{20} +a_{21} =a+18d+a+19d+a+20d$
3a+57d=237
a+19d=79
a=79-19d  ...(2)

Equating (1) and (2),
43-10d=79-19d
9d=36
d=4

Putting the value of d in (1),
a=43-10d
a=43-10(4)
a=43-40
a=3

AP:3, 7, 11, 15, 19, 23...