AP and GP Questions and Answers :
Here we are going to see some practice questions on arithmetic and geometric progression question and answers.
Question 1 :
If the roots of the equation (q − r)x2 + (r − p)x + p − q = 0 are equal, then show that p, q and r are in AP.
If the roots are real, then b2 - 4ac = 0
here, a = q - r, b = r - p and c = p - q
(r - p)2 - 4 (q - r) (p - q)
r2 + p2 - 2rp - 4(pq - q2 - rp + rq) = 0
r2 + p2 - 2rp - 4pq + 4q2 + 4rp - 4rq = 0
r2 + p2 + 2rp - 4pq + 4q2 - 4rq = 0
r2 + p2 + 4q2 + 2rp - 4pq - 4rq = 0
r2 + p2 + (-2q)2 + 2rp - 4pq - 4rq = 0
(r + p - 2q)2 = 0
r + p = 2q
Hence p, q and r in A.P
Question 2 :
If a, b, c are respectively the pth, qth and rth terms of a GP, show that (q − r) log a + (r − p) log b + (p − q) log c = 0.
tn = ARn-1
pth term :
tp = ARp-1 = a
qth term :
tq = ARq-1 = b
rth term :
tr = ARr-1 = c
log a = log A + (p-1) log R -----(1)
log b = log A + (q-1) log R -----(2)
log c = log A + (r-1) log R -----(3)
By applying the values of log a, log b and log c, we get
(q-r) log a + (r-p) log b + (p-q) log c
= (q-r) (log A + (p-1) log R) + (r-p) (log A + (q-1) log R) + (p-q) (log A + (p-1) log R)
= log A [(q - r) + (r - p) + (p - q)] + [(q-r) (p-1) + (r-p) (q-1) + (p-q) (p-1)] log R
= log A (0) + (0 log R)
(q-r) log a + (r-p) log b + (p-q) log c = 0
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