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1.

If `a ,b ,c`are in G.P., then prove that `loga^n ,logb^n ,logc^n`are in A.P.

Answer» Here, `a,b and c` are in G.P.
`:. b^2 = ac`
Taking log both sides,
`log(b^2) = log(ac)`
`=>2logb = loga+logc`
`=>n*2logb = nloga+nlogc`
`=>2logb^n = loga^n+logc^n`
`:. loga^n,logb^n and logc^n` are in A.P.
Discussion
2.

If `a ,b ,c ,d`are in A.P. and `x , y , z`are in G.P., then show that `x^(b-c)doty^(c-a)dotz^(a-b)=1.`

Answer» Let `D` is the common difference in the given A.P.
Then,
`b-a = D, c-b = D, d-c = D`
`:. b-c = -D, c-a = 2D,a-b = -D`
As, `x,y and z` are in G.P.
`:. y^2 = zx`
Now, `x^(b-c)*y^(c-a)*z^(a-b) = x^(-D)*y^(2D)*z^(-D)`
`= (1/(zx))^D*(y^2)^D`
`=(y^2/(zx))^D`
`=((zx)/(zx))^D...[ `As ` y^2 = zx]`
`=1^D = 1`
`:. x^(b-c)*y^(c-a)*z^(a-b) = 1`
Discussion
3.

If `a ,b ,c`are three distinct real numbers in G.P. and `a+b+c=x b ,`then prove that either x < -1 or x > 3

Answer» `a+b+c = xb`
`=>xb-b = a+c`
`=>b(x-1) = a+c`
`=>x-1 = (a+c)/b`
As, `a,b,c` are in G.P,
`:. b^2 = ac => b = sqrt(ac).``:. x-1 = (a+c)/(sqrt(ac))`
`=>x -1 = sqrt(a/c)+sqrt(c/a)`
Now, we know, `m+1/m gt 2` or `m+1/m lt -2.`
`:. sqrt(a/c)+sqrt(c/a) gt 2 or sqrt(a/c)+sqrt(c/a) lt -2.`
`:. x - 1 gt 2 or x-1 lt -2`
`=>x gt 3 or x lt -1.`
Discussion
4.

If `a ,b ,c ,d`are in G.P. prove that:`(a^2+b^2),(b^2+c^2),(c^2+d^2)`are inG.P.`(a^2-b^2),(b^2-c^2),(c^2-d^2)`are inG.P.`1/(a^2+b^2),1/(b^2+c^2),1/(c^2+d^2)`are inG.P.`(a^2+b^2+c^2),(a b+b c+c d),(b^2+c^2+d^2)`

Answer» Let `a` is the first term and `r` is the common ratio.
Then,
`b = ar, c= ar^2,d = ar^3`
`a^2+b^2 = a^2+a^2r^2 = a^2(1+r^2)`
`b^2+c^2 = a^2r^2+a^2r^4 = a^2r^2(1+r^2)`
`c^2+d^2 = a^2r^4+a^2r^6 = a^2r^4(1+r^2)`
So, `a^2+b^2,b^2+c^2 and c^2+d^2` are in G.P. with first term `a^2(1+r^2)` and common ratio `r^2`.
Now, `a^2-b^2 = a^2-a^2r^2 = a^2(1-r^2)`
`b^2-c^2 = a^2r^2-a^2r^4 = a^2r^2(1-r^2)`
`c^2-d^2 = a^2r^4-a^2r^6 = a^2r^4(1-r^2)`
So, `a^2-b^2,b^2-c^2 and c^2-d^2` are in G.P. with first term `a^2(1-r^2)` and common ratio `r^2`.
Now, `1/(a^2+b^2) = 1/(a^2+a^2r^2) = 1/(a^2(1+r^2))`
`1/(b^2+c^2) = 1/(a^2r^2+a^2r^4) = 1/(a^2r^2(1+r^2))`
`1/(c^2+d^2) = 1/(a^2r^4+a^2r^6) = 1/(a^2r^4(1+r^2))`
So, `1/(a^2+b^2),1/(b^2+c^2) and 1/(c^2+d^2)` are in G.P. with first term `1/(a^2(1+r^2))` and common ratio `1/r^2`.
Now, `a^2+b^2+c^2 = a^2+a^2r^2+a^2r^4 = a^2(1+r^2+r^4)`
`ab+bc+cd = a(ar)+ar(ar^2)+ar^2(ar^3) = a^2r(1+r^2+r^4)`
`b^2+c^2+d^2 = a^2r^2 + a^2r^4+a^2r^6 = a^2r^2(1+r^2+r^4)`
So, `a^2+b^2+c^2,ab+bc+cd , b^2+c^2+d^2` are in G.P. with first term `a^2(1+r^2+r^4)` and common ratio `r`.
Discussion
5.

Let `x`be the arithmetic mean and `y ,z`be tow geometric means between any two positive numbers. Then, prove that`(y^3+z^3)/(x y z)=2.`

Answer» Let `a` and `b` are two positive numbers.
Then,
`x = (a+b)/2`
`y^2 = az and z^2 = yb`
`=>y^3 = ayz and z^3 = byz`
Now, `(y^3+z^3)/(xyz) = (ayz+byz)/(((a+b)/2)yz)`
`=2((a+b)yz)/((a+b)yz) =2`
`:. (y^3+z^3)/(xyz) = 2.`
Discussion
6.

If the A.M. of two positive numbers `aa n db(a > b)`is twice their geometric mean. Prove that : `a : b=(2+sqrt(3)):(2-sqrt(3))dot`

Answer» It is given that A.M. is twice of the G.M.
`:. (a+b)/2 = 2sqrt(ab)`
`=>(a+b)/(2sqrt(ab)) = 2/1`
Using componendo and dividendo,
`=>(a+b+2sqrt(ab))/(a+b-2sqrt(ab)) = (3+1)/(3-1)`
`=>(sqrta+sqrtb)^2/(sqrta-sqrtb)^2 = 3/1`
`=>(sqrta+sqrtb)/(sqrta-sqrtb) = sqrt3/1`
Using componendo and dividendo again,
`=>(sqrta+sqrtb+sqrta-sqrtb)/(sqrta+sqrtb-sqrta+sqrtb) = (sqrt3+1)/(sqrt3-1)`
`=>(2sqrta)/(2sqrtb) = (sqrt3+1)/(sqrt3-1)`
`=> sqrta/sqrtb = (sqrt3+1)/(sqrt3-1)`
`=>a/b = (sqrt3+1)^2/(sqrt3-1)^2`
`=>a/b = (3+1+2sqrt3)/(3+1-2sqrt3)`
`=>a/b = (2+sqrt3)/(2-sqrt3)`
`:. a:b = 2+sqrt3 : 2-sqrt3`
Discussion
7.

In a finite G.P. the product of the terms equidistant from thebeginning and the end is always same and equal to the product of first andlast term.

Answer» Let
`a, ar,ar^2,ar^3...ar^(n-1)`is the G.P. with first term as `a` and common ratio `r`.
Then,
`k` th term from the beginning ` = ar^(k-1)`
Now, `k` th term from the end will be `n-k+1` th term from the beginning.
So, `n-k+1` th term from the beginning `= ar^(n-k+1-1) = ar^(n-k)`
Now, product of `k` th term from the beginning and end `= ar^(k-1)*ar^(n-k)`
`= a*ar^(k-1+n-k)`
`=a*ar^(n-1)`
`=` Product of first and last term of the G.P.
Hence, in a finite G.P. the product of the terms equidistant from the beginning and the end is always same and equal to the product of first and last term.
Discussion
8.

Sum the series : `x(x+y)+x^2(x^2+y^2)+x^3(x^3+y^3+ ton`terms.

Answer» `x(x+y)+x^2(x^2+y^2)+....n` terms
`=>x^2+xy+x^4+x^2y^2+x^6+x^3y^3+...n` terms
`=>(x^2+x^4+x^6+...n ` terms `)` +`(xy+(xy)^2+(xy)^3+...n` terms`)`
Now, we have `2` G.P. with first term and common ratio `x^2` and `xy`.
Sum of a G.P. can be given as,
`S_n = (a(r^n-1))/(r-1)`
`:. S_n = (x^2(x^(2n) - 1))/(x^2-1)+(xy(xy^n - 1))/(xy-1)`
Discussion
9.

The `(m+n)t ha n d(m-n)t h`terms fa G.P. ae `pa n dq`respectively. Show that the mth and nth terms are `sqrt(p q)a n dp(q/p)^(m//2n)`respectively.

Answer» Let `a` is the first term of the G.P. is `a` and the common ratio is `r`.
Then, it is given that ,
`ar^(m+n-1) = p->(1)`
`ar^(m-n-1) = q->(2)`
Multiplying (1) and (2),
`a^2r^(m+n-1+m-n-1) = pq`
`=>a^2r^(2(m-1)) = pq`
`=>ar^(m-1) = sqrt(pq)`
`:. m` th term of the given G.P. is `sqrt(pq).`
Now, dividing (1) by (2),
`=>(ar^(m+n-1) )/(ar^(m-n-1) ) = p/q`
`=>r^(2n) = p/q`
`=>r = (p/q)^(1/(2n))`
Now, putting value of `r` in (1),
`=>a((p/q)^(1/(2n)))^(m+n-1) = p`
`=>a = p/((p/q)^((m+n-1)/(2n)))`
`=>ar^(n-1) = p/((p/q)^((m+n-1)/(2n)))**((p/q)^(1/(2n)))^(n-1)`
`=>ar^(n-1) = p(p/q)^(((n-1)/(2n) - (m+n-1)/(2n)))`
`=>ar^(n-1) = p(p/q)^((-m)/(2n))`
`=>ar^(n-1) = p(q/p)^((m)/(2n))`
`:. n` th term of the given G.P. is ` p(q/p)^((m)/(2n)).`
Discussion
10.

In a G.P. of positive terms if any terms is equal to the sum of nexttow terms, find the common ratio of the G.P.

Answer» Let the G.P. is,
`a, ar,ar^2,ar^3...`
Here, `a` is the first term and `r` is the common ratio.
Then, it is given that ,
`ar+ar^2 = a`
`=>r+r^2 = 1`
`=>r^2+r-1 = 0`
`=> r = (-1+-(sqrt(1-4(-1)(1))))/2 =+- (sqrt5-1)/2`
Here, we will take only positive value as it is given that G.P. is of positive terms.
So, common ratio of this G.P. is `(sqrt5-1)/2.`
Discussion
11.

The sum of first two terms of an infinite G.P. is 5 and each term isthree times the sum of the succeeding terms. Find the G.P.

Answer» Let the G.P. is `a, ar,ar^2...oo`
Here, `a` is the first term and `r` is the common ratio.
Then,
`a+ar = 5`
Also,
`T_n = 3(T_(n+1)+T_(n+2)+...oo)`
`=>ar^(n-1) = 3(ar^n+ar^(n+1)+...oo)`
`=>ar^(n-1) =3ar^n(1+r+r^2...oo)`
`=>1 = 3r(1/(1-r))`
`=>1-r = 3r`
`=>4r = 1`
`=> r = 1/4`
`:. a +a(1/4) = 5`
`=>5/4a = 5`
`=> a= 4`
So, G.P. is `4,1,1/4,1/16...oo.`
Discussion
12.

Find the least value of `n`for which the sum `1+3+3^2+ ton`terms is greater than 7000.

Answer» `1+3+3^2+...->n` terms
Here, first term, `a = 1` and common ratio, `r = 3`.
`:. S_n = (1(3^n-1))/(3-1)`
`:. (3^n-1))/(3-1) gt 7000`
`=>3^n -1 gt 14000`
`=>3^n gt 14001`
Now, `3^8 = 6561`
`3^9 = 19683`
So, minimum value of `n` for which `3^n` is greater than `14001` is `9`.
`:. n = 9.`
Discussion