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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The sum of the third and the seventh terms of an ap is 6 and their product is 8 find the sum of first 16 terms of the ap
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tuka81
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Let the first term be a and common difference be d
nth term = a+(n-1)d
Given Third Term + Seventh term = (a+2d)+(a+6d) = 6 ==> a+4d = 3
hence, a= 3-4d
Third Term * Seventh term = (a+2d)*(a+6d) = 8
(3-4d+2d)*(3-4d+6d) = 8==> (3-2d)*(3+2d) = 8
i.e. 9-4d^2 = 8==> d^2 = (9-8)/4 = 0.25==> d = 0.5 or -0.5
Now to check which is correct d...
Substitute and find
Case (a): d= 0.5
a+4d = 3==> a=3-4d = 3-4(0.5)=1
3rd term = a+2d= 1+2*0.5 = 2
7th term = a+6d= 1+6*0.5 = 4
Sum = 6 and Product = 8
Case (b): d= -0.5
a+4d = 3==> a=3-4d = 3-4(-0.5) = 3+2 = 5
3rd term = a+2d= 5+2*(-0.5) = 4
7th term = a+6d= 5+6*(-0.5) = 2
Sum = 6 and Product = 8
Since both are matching, we will go with bothvalues
Sum of first 16 terms = n*(2a+(n-1)d)/2 = 16*(2a+15d)/2
= 8*(2a+15d)
Case (a): d= 0.5
Sum = 8*(2*1+15*0.5)=76
Case (b): d= 0.5
Sum = 8*(2*5+15*(-0.5))=20
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;