Question

evaluate it :-...



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Answer 1

Answer:

Let a be the first term and d be the common difference of the given AP. Then,

S

n

=

2

n

[2a+(n−1)d]

Given,

S

m

=n

2

m

[2a+(m−1)d]=n

2am+m(m−1)d=2n ...........(1)

S

n

=m

2

n

[2a+(n−1)d]=m

2an+n(n−1)d=2m ..........(2)On subtracting 2 from 1, we get,

2a(m−n)+[(m

2

−n

2

)−(m−n)]d=2(n−m)

(m−n)[2a+(m+n−1)d]=2(n−m)

2a+(m+n−1)d=−2 ..........(3)

Sum of (m+n) terms of the given AP

S

m+n

=

2

m+n

[2a+(m+n−1)d]

=

2

m+n

(−2)=−(m+n)

Answer 2

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