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The pth and qth terms of an A.P. are x a...

The pth and qth terms of an A.P. are x and y respectively. Prove that the sum of (p + q) terms is.
`(p+q)/(2)[x+y+(x-y)/(p-q)].`

Text Solution

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Let first term =a and common difference = d
Now, `T_(p)=x rArr a+(p-1)d=x " " ...(1)`
and `T_(q)=y rArr a+(q-1)d=y " " ...(2)`
Subtracting eq. (2) from eq. (1), we get
`(p-q)d=x-y rArr d=(x-y)/(p-q) " " ...(3)`
Now, if you put the value of d in eq. (1) or (2), and find a then it will be a very tedious job.
So, don't find 'a' we will simplify in a special manner :
`S_(p+q)=(p+q)/(2)[2a+(p+q-1)d]`
Bifurcate the terms inside the bracket as
`rArr S_(p+q)=(p+q)/(2)[{a+(p-1)d}+a+(q-1)d}+d]`
`rArr S_(p+q)=(p+q)/(2) [x+y+(x-y)/(p-q)] " " `[from (1),(2) and (3)] `" " ` Hence Proved.
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