Question

eth term of an ap is 1/y and yth term 1/x then it's (xy)th term
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Answer 1

Answer:

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Explanation:

Let a be the first term and d be the common difference of an A.P

Given, xth term is 1/y

→nth term of an A.P is given by a+(n-1)d

Then, xth term will be a+(x-1)d=1/y---(1)

And yth term is 1/x

→ a+(y-1)d=1/x ------(2)

eq(1) -eq(2) gives

(x-1)d-(y-1)d=1/y-1/x

(x-y)d=(x-y)/xy

d=1/xy ------(3)

substituting the value of d in eq(1) ,we get

a+(x-1)(1/xy)=1/y

a+(1/y)-1/xy=1/y

→a=1/xy-----(4)

Now, we know that sum of n terms of an A.P is given by

(n/2)(2a+(n-1)d)

Then sum of xy terms of an A.P will be

(xy/2)(2a+(xy-1)d)

substituting (3) and (4) we get

(xy/2)(2{1/xy}+(xy-1){1/xy})

(xy/2)(2/xy+1-1/xy)

(xy/2)(1+1/xy)