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Question
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.
Solution
`t_n=a(n-1)d`
∴ `t_3=a+(3-1)d=a+2d`
`t_7=a+(3-1)d=a+2d`
∴ `t_3+t_7=(a+2d)+(a+6d)=2a+8d`
∴ `2a+8d=6`
∴ a+4d=3 ................(I)
`t_3xxt_7=(a+2d)(a+6d)`
= `(a+4d-2d) (a+4d+2d)`
=`(3-2d) (3+2d) `......................from (I)
∴ `(3-2d)(3+2d)=8`
∴ `9-4d^2=8`
∴` 4d^2=1 d^2=1/4 d=1/2 or d=-1/2`
Now, `if d=1/2`
`a+4xx1/2=3`................ from (I)
`a=1 `
If ` d=-1/2`
`a+4xx(-1/2)=3.......` from (I)
`a=5`
∴ the first term of the A. P. is 1 and the common difference is `1/2.`
or , the first term of the A. P. is 5 and the common difference is `-1/2`.
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