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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.
Step-by-step explanation:
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
= 2/1
[2×5+(16−1)×(− 2/1 )]=8(10− 215 )=4×5=20
Thus, the sum of first 16 terms of the AP is 76 or 20.
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