2. The sum of the third and the seventh terms
of an AP is 6 and their product is 8. Find
the sum of first sixteen terms of the AP.
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Answer:
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The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P
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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.
Explanation:
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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.