The sum of the third and seventh terms of an AP is 6 and their product is 8 . Find the sum of first sixteen terms of an AP.
Answers
Answer:
Let a and d be the first term and common difference of AP
nth term of AP
a
n
=a+(n−1)d
∴a
3
=a+(3−1)d=a+2d
a
7
=a+(7−1)d=a+6d
Given a
3
+a
7
=6
∴(a+2d)+(a+6d)=6
⇒2a+8d=6
⇒a+4d=3....(1)
Also given
a
3
×a
7
=8
∴(a+2d)(a+6d)=8
⇒(3−4d+2d)(3−4d+6d)=8 [Using (1)]
⇒(3−2d)(3+2d)=8
⇒9−4d
2
=8
⇒4d
2
=1
⇒d
2
=
4
1
⇒d=±
2
1
When d=
2
1
a=3−4d=3−4×
2
1
=3−2=1
When d=−
2
1
a=3−4d=3+4×
2
1
=3+2=5
When a=1 & d=
2
1
S
16
=
2
16
[2×1+(16−1)×
2
1
]=8(2+
2
15
)=4×19=76
When a=5 & d=−
2
1
S
16
=
2
16
[2×5+(16−1)×(−
2
1
)]=8(10−
2
15
)=4×5=20
Thus, the sum of first 16 terms of the AP is 76 or 20.
How satisfied are you with the answer?
This will help us to improve better
ASK
Get Instant Solutions, 24x7
No Signup required
girl
Solved Examples
EASY
MEDIUM
HARD
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;