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)x^{2} + (r − p)x + p − q = 0 are equal, then show that p, q and r are in AP.
Solution :
If the roots are real, then b^{2} - 4ac = 0
here, a = q - r, b = r - p and c = p - q
(r - p)^{2} - 4 (q - r) (p - q)
r^{2} + p^{2} - 2rp - 4(pq - q^{2} - rp + rq) = 0
r^{2} + p^{2} - 2rp - 4pq + 4q^{2} + 4rp - 4rq = 0
r^{2} + p^{2} + 2rp - 4pq + 4q^{2} - 4rq = 0
r^{2} + p^{2} + 4q^{2 }+ 2rp - 4pq - 4rq = 0
r^{2} + p^{2} + (-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.
Solution :
t_{n } = AR^{n-1}
p^{th} term : t_{p } = AR^{p-1 } = a |
q^{th} term : t_{q } = AR^{q-1 } = b |
r^{th} term : t_{r } = AR^{r-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
Hence proved.
After having gone through the stuff given above, we hope that the students would have understood "Arithmetic and Geometric Progression Question and Answers".
Apart from the stuff given above, if you want to know more about "Arithmetic and Geometric Progression Question and Answers". Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
May 22, 24 06:32 AM
May 17, 24 08:12 AM
May 14, 24 08:53 AM