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Answer:
178
Step-by-step explanation:
Tn=a+(n-1)d
T5=38 and T9=66
T5=a+(5-1)d T9=a+(9-1)d
38=a+4d-------->eq 1. 66=a+8d---------->eq 2
By simultaneous equation method
d=7 and a=10
Then by the formula of Tn=a+(n-1)d
T25=10+(25-1)7
=10+(24)7
=10+168
=178
Answer :
a(25) = 178
Note :
★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.
★ If a1 , a2 , a3 , . . . , an are in AP , then
a2 - a1 = a3 - a2 = a4 - a3 = . . .
★ The common difference of an AP is given by ; d = a(n) - a(n-1) .
★ The nth term of an AP is given by ;
a(n) = a + (n - 1)d .
★ If a , b , c are in AP , then 2b = a + c .
★ The sum of nth terms of an AP is given by ; S(n) = (n/2)×[ 2a + (n - 1)d ] .
or S(n) = (n/2)×(a + l) , l is the last term .
★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .
Solution :
→ Given :
• 5th term , a(5) = 38
• 9th term , a(9) = 66
→ To find :
• 25th term , a(25) = ?
Now ,
We know that , the nth term of an AP is given as ; a(n) = a + (n - 1)d .
Also ,
We have , 5th term = 38
=> a(5) = 38
=> a + (5 - 1)d = 38
=> a + 4d = 38 ------------(1)
Also ,
9th term = 66
=> a(9) = 66
=> a + (9 - 1)d = 66
=> a + 8d = 66 -----------(2)
Now ,
Subtracting eq-(1) from eq-(2) , we have ;
=> (a + 8d) - (a + 4d) = 66 - 38
=> a + 8d - a - 4d = 28
=> 4d = 28
=> d = 28/4
=> d = 7
Now ,
Putting d = 7 in eq-(1) , we have ;
=> a + 4d = 38
=> a + 4×7 = 38
=> a + 28 = 38
=> a = 38 - 28
=> a = 10
Now ,
Using the formula a(n) = a + (n - 1)d ;
The 25th term of the AP will be given as ;
=> a(25) = a + (25 - 1)d
=> a(25) = a + 24d
=> a(25) = 10 + 24×7
=> a(25) = 10 + 168
=> a(25) = 178
Hence , a(25) = 178 .