in an ap, if pth term is 1/q and qth term is 1/p profe thatthe sum of fist pq term is 1+pq/2 where p≠q
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let a is the first term and d is the common difference of AP
now,
pth term =1/q
a+(n-1) d=1/q ---------(1)
again
qth term=1/p
a+(q-1) d=1/p -------------(2)
solve equation (1) and (2)
d=1/pq
a =1/pq
now ,
sum of (pq)th term =pq/2 {2/pq+(pq-1) 1/pq}
=pq/2 {(1+pq)/pq}
= (1+pq)/2
hence proved
now,
pth term =1/q
a+(n-1) d=1/q ---------(1)
again
qth term=1/p
a+(q-1) d=1/p -------------(2)
solve equation (1) and (2)
d=1/pq
a =1/pq
now ,
sum of (pq)th term =pq/2 {2/pq+(pq-1) 1/pq}
=pq/2 {(1+pq)/pq}
= (1+pq)/2
hence proved
abhi178:
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Answer:
Step-by-step explanation:
yes he or she is right
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