Question 508360: if 12th and 25th term of an airthmetic progression are 40 and 131 .then what is the 50th term
Answer by Maths68(1474)   (Show Source):
You can put this solution on YOUR website! Use
Tn=a+(n-1)d
Where
Tn = Term
n = Term number
a = First term
d = Common Difference
============================================================
For 12th Term which is 40
Tn = 40, n = 12, a = ?, d = ?
Tn=a+(n-1)d
Put the values in above equation
40=a+(12-1)d
40=a+(11)d
40=a+11d.............(1)
===============================================================
For 25th Term which is 131
Tn = 131, n = 25, a = ?, d = ?
Tn=a+(n-1)d
Put the values in above equation
131=a+(25-1)d
131=a+(24)d
131=a+24d..............(2)
==============================================================
Now we have two simultaneous equations and two unknowns (a and d)
Subtract (1) from (2)
131=a+24d..............(2)
-40=-a-11d.............(1)
--------------
91=13d
91/13=d
7=d
Put the value of d in (1)
40=a+11d.............(1)
40=a+11(7)
40=a+11(7)
40=a+77
40-77=a
-37=a
For 50th term
Tn=?, n=50, a=-37, d=7
Again use
Tn=a+(n-1)d
Tn=-37+(50-1)7
Tn=-37+(49)7
Tn=-37+343
Tn=306
the 50th term of given A.P. is 306
A.P will be
-37,-30,-23,16,9,2,5,12,19,26,33,40,47,54,...............
|
|
|
