**Question 11. **If the GP’s 5, 10, 20,…….. and 1280, 640, 320,……..have their nth terms equal, find the value of n.

**Solution:**

For GP 5, 10, 20,…………..

a1 = 5,

r = a2/a1 =10/5 = 2

For GP 1280, 640, 320,…………..

a1′ = 1280,

r’ = a2’/a1′ =640/1280 = 1/2

Given nth terms are equal

Hence, a1*r

^{n-1}= a1’*r’^{n-1}5*2

^{n-1 }= 1280*(1/2)^{n-1}2

^{n-1}* 2^{n-1}= 1280/52

^{2n-2}= 2562

^{2n-2}= (2)^{8}2n-2 = 8

2n = 10

n = 5

**Question 12.** If the 5th, 8th and 11th terms of a GP are p, q and s respectively. Prove that q^{2} = ps.

**Solution:**

Let a1 be first term and common ratio be r of given GP

Given a5 = p, a8 = q, a11 = s

a8/a5 = q/p

(a1*r

^{7})/(a1*r^{4}) = q/pr

^{3}= q/p –Equation Ia11/a8 = s/q

(a1*r

^{10})/(a1*r^{7}) = s/qr

^{3}= s/q — Equation IIFrom Equation, I and Equation II

q/p = s/q

q

^{2}= psHence, proved that q

^{2 }= ps.

**Question 13. **The 4th term of a GP is square of its 2nd term and the first term is -3. Find its 7th term.

**Solution:**

Let a1 be first term and common ratio be r of given GP

Given a4 = (a2)

^{2}, a1 = -3a1*r3 = (a1*r)

^{2}-3*r3 = (-3*r)

^{2}-3*r

^{3}= 9*r^{2}r = -3

7th term of given GP is given by

a7 = a1*r6

= -3*(-3)

^{6}= (-3)

^{7}= -2187

**Question 14.** In GP the 3rd term is 24 and the 6th term is 192. Find the 10th term.

**Solution:**

Let a1 be first term and common ratio be r of given GP

Given a3 = 24, a6 = 192

a6/a3 = 8

(a1*r

^{5})/(a1*r^{2}) = 8r

^{3}= 8r = 2

We have a3=24

a1*r

^{2}= 24a1*2

^{2}= 24a1*4 = 24

a1 = 6

The 10th term of given GP is given by

a10 = a1*r

^{9}

^{ }= 6*2^{9}= 3072.

**Question 15. **If a, b, c, d and p are different real numbers such that:

**(a ^{2}+b^{2}+c^{2})p^{2} – 2(ab+bc+cd)p + (b^{2}+c^{2}+d^{2}) ≤ 0, then show that a, b, c and d are in GP.**

**Solution:**

Given, (a

^{2}+b^{2}+c^{2})p^{2}– 2(ab+bc+cd)p + (b^{2}+c^{2}+d^{2}) ≤ 0(a

^{2}p^{2}+b^{2}p^{2}+c^{2}p^{2}) – 2(abp+bcp+cdp) + (b^{2}+c^{2}+d^{2}) ≤ 0(a

^{2}p^{2}+b^{2}-2abp) + (b^{2}p^{2}+c^{2}-2bcp) + (c^{2}p^{2}+d^{2}-2cdp) ≤ 0(ap-b)

^{2}+ (bp-c)^{2}+ (cp-d)^{2 }≤ 0Sum of squares cannot be less than 0.

Hence, (ap-b)

^{2}+ (bp-c)^{2}+ (cp-d)^{2}= 0(ap-b)

^{2}= 0p = b/a

(bp-c)

^{2}= 0p = c/b

(cp-d)

^{2}=0p = d/c

b/a = c/b = d/c

Hence, a, b, c and d are in GP.

**Question 16. **If (a+bx)/(a-bx) = (b+cx)/(b-cx) = (c+dx)/(c-dx) (x≠0), then show that a, b, c and d are in GP.

**Solution:**

Given (a+bx)/(a-bx) = (b+cx)/(b-cx)

Applying componendo and dividendo

(a+bx)+(a-bx) / (a+bx)-(a-bx) = (b+cx)+(b-cx) / (b+cx)-(b-cx)

2a/2bx = 2b/2cx

a/b =b/c –Equation I

Similarly, (b+cx)/(b-cx) = (c+dx)/(c-dx)

Applying componendo and dividendo

(b+cx)+(b-cx) / (b+cx)-(b-cx) = (c+dx)+(c-dx) / (c+dx)-(c-dx)

2b/2cx = 2c/2dx

b/c =c/d –Equation II

From Equation, I and Equation II

a/b = b/c = c/d

b/a = c/b = d/c

Hence, a, b, c and d are in GP.

**Question 17.** If the pth and qth terms of a GP are q and p respectively. Show that (p+q)th term is (qp/pq)1/p-q.

**Solution:**

Let the first term of given GP be a1 and common ratio be r.

Given, ap = q and aq = p

aq/ap = p/q

(a1*r

^{q-1})/(a1*r^{p-1}) = p/qr

^{(q-p)}= p/qr = (p/q)

^{1/(q-p)}ap = q

a1*r

^{p-1}= qa1*(p/q)

^{(p-1)/(q-p)}= qa1 = ((q/p)

^{(p-1)/(q-p)})* qThe (p+q)th term of given GP is given by

a(p+q) = a1*r

^{(p+q-1)}= [((q/p)

^{(p-1)/(q-p)})* q]*[(p/q)^{1/(q-p)}]^{(p+q-1)}= [((q/p)

^{(p-1)/(q-p)})* q]*[(p/q)^{(p+q-1)/(q-p)}]= q*[(q/p)

^{(p-1)/(q-p)}]*[(p/q)^{(p+q-1)/(q-p)}]= q*[(p/q)

^{((1-p)+(p+q-1))/(q-p) }]= q*[(p/q)

^{q/(q-p)}]= q*[(q/p)

^{q/(p-q)}]= [q

^{p}/p^{q}]^{1/(p-q)}