The sum of third and seventh term of an AP is 6 and their product is 8.Find the sum of the first sixteen terms of the AP.
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Step-by-step explanation:
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5 positive integers are in AP. the sum of three middle term is 24. product of first and 5th term is 48.find the terms of AP
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Step-by-step explanation:
Let the 5 +ve integers be
a-3d,a-d,a,a+d,a+3d
sum of middle 3 No's
a-d+a+a+d=24
3a=24
=>a=8
product of first n last term
(a-3d)×(a+3d)=48
a^2-(3d)^2=48
(8^2) - (3^2)(d^2) = 48
64-9d^2=48
-9d^2=48-64
-9d^2 = -16
d^2=16/9
d=4/3
so the required terms of the AP are,,
a-3d=8-3(4/3)=4
a-d=8-4/3=20/3
a=8
a+d=8+4/3=28/3
a+3d=8+3(4/3)=12
therefore, the AP is......
4, 20/3, 8, 28/3, 12
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;