If p times the pth term of an AP is an equal to q times the q th term. Find the (p+q)th term of the AP
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let a is the first term and d is the common difference of an AP
so,
a/ c to question,
P×Tp = q× Tq
P×{ a + ( P -1)d} = q×{ a + ( q -1)d}
Pa + P(P -1)d = qa + q(q -1)d
(P-q )a = d{ q² -q -p² +p}
(P-q)a = d{ (q -P)(q + P) -(q -p) }
(p -q)a = -(p-q)d {P+ q - 1}
a + ( p +q -1)d = 0 ----------(1)
now,
T(p + q) = a + (P+q -1)d
from equation (1)
T(P +q) = 0
so,
a/ c to question,
P×Tp = q× Tq
P×{ a + ( P -1)d} = q×{ a + ( q -1)d}
Pa + P(P -1)d = qa + q(q -1)d
(P-q )a = d{ q² -q -p² +p}
(P-q)a = d{ (q -P)(q + P) -(q -p) }
(p -q)a = -(p-q)d {P+ q - 1}
a + ( p +q -1)d = 0 ----------(1)
now,
T(p + q) = a + (P+q -1)d
from equation (1)
T(P +q) = 0
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