Question

Answer it fast with step by step plz. solve both of them

Answer 1

LET 1st term be a
the common difference be d
S1=n/2(2a+(n-1)d)->I

S2=2n/2(2a+(2n-1)d)->II

S3=3n/2(2a+(3n-1)d)->III
we have to prove :
S3= 3 (S2-S1)
solve RHS side
= 3(2n/2 (2a+ (2n-1) d) -n/2 (2a+ (n-1) d
= 3*n/2 (4a+4n-2) d-2a-(n-1)d
= 3n/2 (2a+4n-2-n+1)d
= 3n/2 (2a+(3n-1)d



S3=3(S2-S1)

proove that


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Given that :-

an = p

Prove that : S2n-1 = (2n-1)p

{(2n-1)/2}*[2a+(2n-1-1)d] = S2n-1

{(2n-1)/2}*[2a+(2n-2)d] = S2n-1

(2n-1)* [ a + (n-1)d ] = S2n-1

Given that :- ( a + (n-1)d ) = p

then

(2n-1)p = S2n-1