Solution:-
Given that
10th term is 31
Means ,a10=31
a+9d=31........ 1
and 15th term is 66
Means ,a15=66
a+14d=66...... (1) and (2)
a+9d=31
a+14d=66
￣￣￣￣￣
After solving , a=-32 , d=7
31st term
Means, a31
=a+29d
=-32+29×7
=-32+203
=171 ans.
the nth term of an ap is given by an=3+4n the common difference is
Find the first 10 terms of the sequence an 4n+3
Q.Find 25th term of an AP whose 10th term is 38 and 14th term is 43.
Solution:-
Given that
10th term is 38
Means ,a10=38
a+9d=38........ (1)
and 14th term is 43
Means ,a14=43
a+13d=43...... (2)
From (1) and (2)
a+9d=38
a+13d=43
￣￣￣￣￣
After solving , a=107/4 , d=5/4
25 st term
Means, a25
=a+24d
=107/4+24×5/4
=227/4 ans...
Q.Second term of an arithmetic sequence is 5 and fifth term is 2 The common differnce will be
solution:- Given that a2 =5
a5 =2
now,a2 =5
a+d=5......(1)
a5 =2
a+4d=2.....(2)
From equation (1) and (2)
a+d=5
a+4d=2
after solving the equation
d=-1 and a=6
hence the common difference -1. ans....
