**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 t |
q t |
r t |

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".

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