NCERT Class XI Mathematics - Sequences and Series - Solutions
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Question : 93
Total: 106
The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that (q – r) a + (r – p) b + (p – q) c = 0
Solution:
Let A be the first term and D be the common difference of A.P.,
then
a p
a q
a r
Considering (q – r) a + (r – p) b + (p – q) c
= (q – r) [A + (p – 1) D] + (r – p) [A + (q – 1) D] + (p – q) [A + (r – 1) D]
= A [(q – r) + (r – p) + (p – q)] + D[(q – r)(p – 1) + (r – p)(q – 1) + (p – q)(r – 1)]
= A[0] + D[0] = 0
Hence, (q – r) a + (r – p)b + (p – q) c = 0.
then
Considering (q – r) a + (r – p) b + (p – q) c
= (q – r) [A + (p – 1) D] + (r – p) [A + (q – 1) D] + (p – q) [A + (r – 1) D]
= A [(q – r) + (r – p) + (p – q)] + D[(q – r)(p – 1) + (r – p)(q – 1) + (p – q)(r – 1)]
= A[0] + D[0] = 0
Hence, (q – r) a + (r – p)b + (p – q) c = 0.
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