The sum of third and seventh term of an A. P. is 6 and their product is
8. Find the first term and the common difference of the A. P.
Answers
A3 + A7= 6 ( given)
a+ 2d+a+6d= 6
2a+ 8d= 6
2( a+4d)=6
a+4d=3
a=3-4d.......(i)
A3×A7=8
a+2d)(a+6d)=8.....(ii)
putting the value of a in equation (ii)
(3-4d+2d)(3-4d+6d)=8
(3-2d)(3+2d)=8
(3)^2- (2d)^2=8
9-4d^2=8
9-8= 4d^2
1= 4d^2
1/4=d^2
so,,,d= 1/2
putting the value of a in eq. ( i)
a=3-4d
a= 3-4×1/2
a= 1
Answer:
So using formula for n th term
a = first term
n = no. of terms
d = common difference
So given
a + (3-1)d + a (7-1)d = 6
2a + 8d = 6
a + 4d = 3
a = 3 - 4d
(a + 2d)(a + 6d) = 8
(3 - 2d)( 3 + 2d) = 8 {substituting a = 3 - 2d}
9 - 4d² = 8
d = +1/2 and - 1/2
so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2 a = 3 + 2 = 5
so using formula for
when d = +1/2
Similarly when d = -1/2
4.6
523 votes
THANKS
859
Comments Report
SachinKrMb Genius
Given :-
→ a₃ + a₇ = 6.
→ a₃ × a₇ = 8 .
To find :-
→ S₁₅ .
Solution :-
We have,
→ a₃ + a₇ = 6.
⇒ a + 2d + a + 6d = 6 .
⇒ 2a + 8d = 6 .
⇒ 2( a + 4d ) = 6 .
⇒ a + 4d = 6/2 .
⇒ a + 4d = 3
∵a = 3 - 4d ............(1) .
And,
→ a₃ × a₇ = 8 .
⇒ ( a + 2d ) × ( a + 6d ) = 8 .
⇒ ( 3 - 4d + 2d ) ( 3 - 4d + 6d ) = 8 .
⇒ ( 3 - 2d )( 3 + 2d ) = 8 .
⇒ 3² - (2d)² = 8 .
⇒ 9 - 4d² = 8.
⇒ 4d² = 9 - 8 .
⇒ 4d² = 1 .
⇒ d² = 1/4 .
⇒ d = √(1/4) .
∴ d = 1/2 .
Putting the value of d in equation (1), we get ,
⇒ a = 3 - 4 × 1/2 .
⇒ a = 3 - 2 .
∴ a = 1 .
s16=76