If the 7th term of an AP is 20 and 13th term is 32, find the AP and its 25th term.
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Answer:
Step-by-step explanation:
We have ,
t7= 20
t13= 32
t25= ?
tn = a +(n-1)×d
t7= a+ 6d------ equation 1
Similarly,
t13= a+ 12d------ equation 2
Now, subtracting eq 1 from eq 2
We get ,. 6d= 12
d= 12/6=2
Hence , common difference is 2
Put d=2 in eq 1 we get ,
a+6×2=20
a=20-12
a=8
Therefore first term (a) = 8
Now,
t1=a=8
t2=t1+d=8+2=10
t3=t2+d=10+2=12
t4=t3+d=12+2=14
Therefore the sequence forms in AP as 8,10,12,14,.... With d=2
Now, to find 25^th term
tn= a+(n-1)×d
t25= 8+(25-1)×2
= 8+24×2
= 8+48
= 56
Hence t25 = 56
Ans: First term is 8
Common difference is 2
The sequence forms in AP as 8,10,12,14,...
t25= 56
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