Algebra theory class - Sequences continued

Q1. Interior angle of convex polygon are in AP with smallest angle being 120° & common difference 5°. Find no. of sides.

Solution:
Sum of internal angles of n-sided polygon
= (n-2)*180
= n/2[2a + (n-1)*5]
Once you simplify this, you will get this quadratic:
(n-16)(n-9) = 0
=> n = 9,16
But each internal angle has to be less than 180.
It will hold only with n = 9 and not n = 16.
Answer n = 9.

Proof for Sum of internal angles of n-sided polygon
e1 + e2 .. en = 360
e1 + i1 + e2 + i2 ... + en + in = n*180
=> 360 + i1 + i2 ... in = n*180
=> sum of internal angles = 180(n-2)

= (n-2)*180

Q2. An infinite GP has sum 2005. A new infinite GP is formed with squaring every term such that new sum is 10 times of previous sum. Find common ratio of original series.

Solution:
a/(1-r) = 2005__________[1]
a^2/(1-r^2) = 2005*10______[2]
[2]/[1]^2
=> (1-r)^2/(1-r^2) = 10/2005
=> (1-r)/(1+r) = 10/2005 = 2/401

By Componendo/Dividendo:
a/b = c/d => a+b/a-b = c+d/c-d
Why?
a/b + 1 = c/d + 1 => a+b/b = c+d/d
Similarly:
a-b/b = c-d/d
a-b/c-d = b/d = a+b/c+d
=> a+b/a-b = c+d/c-d

So 1+r/1-r = 401/2 => 2r/2 = 399/403 = r = answer

Q3. 
Sum of A.P. is 715. First term is increased by 1, second term is increased by 3, third by 5 & so on.
Sum of new sequence is 836. Find sum of first term, middle term & last term of original AP.

Solution:
O.A.P = a1 + a2 ... an = 715
1 + 3 ... (2n-1) = 836 - 715 = n/2[2 + 2n - 2] = n^2 = 121 => n = 11

Now let's rewrite first A.P.:
a1 + a11 = 2a1 + 10d
a2 + a10 = 2a1 + 10d
a3 + a9 = 2a1 + 10d
a4 + a8 = 2a1 + 10d
a5 + a7 = 2a1 + 10d
a6 = a1 + 5d
715 = 11a1 + 55d = 11(a1 + 5d)
a1 + a6 + a11 = 3(a1 + 5d) = 715*3/11 = 65*3 = 195

Q4. 
There are even no. of terms in A.P. such that sum of odd position terms is 24 & sum of even position terms is 30. Also difference of last & first term is 10.5. Find number of terms in AP.

Solution:
a1 + (a1 + 2d) + (a1 + 4d) ... (a1 + (2n-2)d) = 24_____[1]
(a1+d) + (a1 + 3d) + (a1 + 5d) ... (a1 + (2n-1)d) = 30_______[2]
[2] - [1]
=> 6 = n*d____[3]
Now,
10.5 = (2n-1)*d = 2nd - d = 12 - d => d = 1.5 => n = 6/1.5 = 4.
n = 4 => 2n = 8 = total number of terms = Answer.

Q5.
4 positive increasing integers are there such that first 3 are in AP & last 3 are in GP.
Also difference of last & first is 30.
Find sum of four terms.

Solution:
a, a+d, a+2d,k
k*(a+d) = (a+2d)^2
=> k = (a+2d)^2/(a+d)
(a+2d)^2/(a+d) - a = 30
=> a^2 + 4ad + 4d^2 - a^2 - ad = 30(a+d)
=> 3ad + 4d^2 = 30(a+d)
=> 4d^2 - 30d = 30a - 3ad
=> 2d(2d - 15) = 3a(10-d)
Since a,d are positive (as given in the question) sign(2d - 15) = sign(10-d)
=> (2d-15)(10-d) > 0
=> (d-7.5)(d-10) < 0
=> 7.5 < d < 10
But d is integer => d = 8,9
d = 8 => a = 8/3 so d = 9 should be it.
Answer: d = 9

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