Question

Find an A.P. whose fourth term is 9 and the sum of its 6th term and 13th term is 40.

Answer 1

Given:-

a4=9......i
a6+a13=40......ii

From i

a4=9
a+(4-1)d=9
a+3d=9.........iii×2

From ii

a6+a13=40

a+5d+a+12d=40
2a+17d=40...........iv

Sub (iii)×2 from iv

2a+6d=18
2a+17d=40
(-) (-) (-)
__________
-11d=-22

d=-22/-11
d=2

putting value of d in equation i
a+3d=9
a+3×2=9
a+6=9
a=9-6
a=3


to form AP..
a,a+d,a+2d..........

3,3+2,3+2×2....
3,5,3+4.....
>3,5,7..........is the required AP.


@Altaf