In an ap xth term is 1/y and yth term is 1/x . Find the sum of the first xy terms
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Hi friend!!
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)
I hope this helps!
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)
I hope this helps!
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