5. The sum of the 3rd and the 7th terms of an
A.P. is 6 and their product is 8. Find the first
term and the common difference.
Answers
Answer:
Let's consider that the AP in consideration has ;
First term = a
Common difference = d
Also, we know that ,
The nth term of an AP is given by;
T(n) = a + (n - 1)d
Thus,
=> T(3) = a + (3 - 1)d
=> T(3) = a + 2d
Also,
=> T(7) = a + (7 - 1)d
=> T(7) = a + 6d
Now, according to the question,
The sum of the 3rd and 7th terms of the AP is 6.
=> T(3) + T(7) = 6
=> a + 2d + a + 6d = 6
=> 2a + 8d = 6
=> 2(a + 4d) = 6
=> a + 4d = 6/2
=> a + 4d = 3
=> a = 3 - 4d
Also , it is given that ,
Product of the 3rd and 7th terms is 8.
=> T(3)•T(7) = 8
=> (a + 2d)(a + 6d) = 8
Putting a = 3 - 4d , we get;
=> (3 - 4d + 2d)(3 - 4d + 6d) = 8
=> (3 - 2d)(3 + 2d) = 8
=> 3^2 - (2d)^2 = 8
=> 9 - 4d^2 = 8
=> 4d^2 = 9 - 8
=> d^2 = 1/4
=> d = ±√(1/4)
=> d = ± 1/2
Thus, using eq-(1) , we get;
=> a = 3 - 4d
Case(1)
when, d = 1/2
=> a = 3 - 4•(1/2)
=> a = 3 - 2
=> a = 1
Case (2)
when, d = -1/2
=> a = 3 - 4•(-1/2)
=> a = 3 + 2
=> a = 5.
Thus , there will be two APs corresponding to different values of a and d .
First AP : a = 1 , d = 1/2
Second AP : a = 5 , d = -1/2
Verification :
For first AP: a = 1 , d = 1/2
T(3) = a + 2d
= 1 + 2•(1/2)
= 1 + 1
= 2
T(7) = a + 6d
= 1 + 6•(1/2)
= 1 + 3
= 4
Sum = T(3) + T(7)
= 2 + 4
= 6
Product = T(3)•T(7)
= 2•4
= 8
For second AP: a = 5 , d = -1/2
T(3) = a + 2d
= 5 + 2•(-1/2)
= 5 - 1
= 4
T(7) = a + 6d
= 5 + 6•(-1/2)
= 5 - 3
= 2
Sum = T(3) + T(7)
= 4 + 2
= 6
Product = T(3)•T(7)
= 4•2
= 8