the sum of third and seventh terms of AP is 6 and their product is 8. find the sum of first sixteen terms of the AP
Answers
a+2d+a+6d=6
2a+8d=6
a+4d=3
The fifth term is 3
fifth term - 2d = third term
fifth term +2d = seventh term
(3-2d)(3+2d)=8
3^2-2d^2=8
9-4d^2=8
9-8=4d^2
1=4d^2
On square rooting
1=2d
d=1/2
a+4d=3
a+4(1/2)=3
a+2=3
a=1
Sixteenth term is a+15d
1+7.5=8.5
Sum of n terms=n/2(a+n^th term)
sum of sixteenth term = 16/2(1+8.5)
8(9.5)=76 ans
Answer :-
a3 + a7 = 6 …………………………….(i)
And
a3 × a7 = 8 ……………………………..(ii)
By the nth term formula,
an = a + (n − 1)d
Third term, a3 = a + (3 -1)d
a3 = a + 2d………………………………(iii)
And Seventh term, a7 = a + (7 -1)d
a7 = a + 6d ………………………………..(iv)
From equation (iii) and (iv), putting in equation(i), we get,
a + 2d + a + 6d = 6
2a + 8d = 6
a+4d=3
or
a = 3 – 4d …………………………………(v)
Again putting the eq. (iii) and (iv), in eq. (ii), we get,
(a + 2d) × (a + 6d) = 8
Putting the value of a from equation (v), we get,
(3 – 4d + 2d) × (3 – 4d + 6d) = 8
(3 – 2d) × (3 + 2d) = 8
3^2 – 2d^2 = 8
9 – 4d2 = 8
4d2 = 1
d = 1/2 or -1/2
Now, by putting both the values of d, we get,
a = 3 – 4d = 3 – 4(1/2) = 3 – 2 = 1, when d = ½
a = 3 – 4d = 3 – 4(-1/2) = 3+2 = 5, when d = -1/2
We know, the sum of nth term of AP is;
Sn = n/2 [2a + (n – 1)d]
So, when a = 1 and d=1/2
Then, the sum of first 16 terms are;
S16 = 16/2 [2 + (16 – 1)1/2] = 8(2+15/2) = 76
And when a = 5 and d= -1/2
Then, the sum of first 16 terms are;