Quiz 4

The sum of the first 25 terms, S₂₅ = 525
The sum of the next 25 terms, K₂₅ = sum of next 25 terms = 725 
a₂₆ = a₁ + 25d
a₂₇ = a₂ + 25d

K₂₅ = a₂₆ + a₂₇ + ...... + a₅₀
Or K₂₅ = a₁ + 25d + a₂ + 25d + ...... + a₂₅ + 25d
Or K₂₅ = a₁ + a₂ + ...... + a₂₅ + 25(25d)

Or K₂₅ = S₂₅ + 25(25d)
i.e., 725 = 525 + 25 * 25d
200 = 25 * 25d 
8 = 25d
d = 825825

The question is "what is the common difference?"

Hence the answer is "8/25"

I. c is the arithmetic mean of a and b => c lies in between a and b. And it lies exactly in between the two terms. As in the number of terms between a and c should be equal to number of terms between b and c.

a, c, b could be the
1st 2nd and 3rd terms respectively or
1st, 3rd and 5th respectively, or
2nd, 3rd, 4th respectively, or
3rd, 4th, 5th respectively.

The terms could also be the other way around. As in, b, c, a could be the 1st 2nd and 3rd terms respectively, or the 1st, 3rd and 5th respectively, and so on. This is a very simple but very powerful idea.

II. Now, d is the arithmetic mean of b and c. => d lies between b and c. Using statements I and II we can say that a, c, b have to be 1st, 3rd and 5th or 5th, 3rd and 1st as there is an element between b and c also.

So, c is the third term. a and b are 1st and 5th in some order.

b ___ c ___ a or a ___ c ___ b

d is the arithmetic mean of b and c.

Possible arrangements are:

b     d   c   e   a    or   a   e   c    d   b

The question is \"Which of the given statements are true?\"

Hence the answer is \"i and ii only\"

Choice A is the correct answer.

Median is the first term => Common ratio has to be negative. why?
Let us see why this is true.
When a > 0,
If r > 1, this will be an increasing G.P.
If r lies between (0, 1), this will be a decreasing G.P.

In both cases, the middle term will be the median. If a < 0, the order will be the other way around, but the middle term would still be the median.
If the middle term is not the median, we can say that r < 0. Now, let us go the solution
b = 10, a and c should be negative. Solution, a + b + c cannot be 15.
a + b + c = –15
10r10r + 10 + 10r = -15 ⇒⇒ 2r2r + 2r = -5
Solving the quadratic, we will get r = −12−12 or -2.
The sequence is either – 5, 10, – 20 or – 20, 10, – 5.
a > c ==> the sequence has to be – 5, 10, – 20.
The product of the first 4 terms = – 5 * 10 * –20 * 40 = 40000.

The question is "what is the product of the first 4 terms of this G.P.?"

Hence the answer is "40,000"

4t₄ = 9t₉ , we need to find t₁₃
4(a + 3d) = 9(a + 8d)
4a + 12d = 9a + 72d
=> 5a + 60d = 0
=> a + 12d = 0
=> t₁₃ = 0
=> 13 * t₁₃ = 0
As a simple rule, if n * t_n = m * t_m, then t_m+n = 0. See, if you can prove this.

The question is "what is 13 times the 13th term of this A.P.?"

Hence the answer is "0"

Sequence P is an A.P. with a = 11, and common difference 3.
So, P_k = 11 + (k – 1)3.
Sequence Q is an A.P. with third term 103 and common difference – 4.
t₃ = a + 2d
103 = a + 2 (– 4) or a = 111
q_k+2 = 111 + (k +1) (– 4)
q_{k + 2} = 111 – 4k – 4 = 107 – 4k
p_k > q_{k + 2}
11 + (k–1)3 > 107 – 4k
8 + 3k > 107 – 4k
7k > 99
k > 997997
k has to be an integer, so smallest value k can take is 15.

The question is "what is the smallest value k can take?"

Hence the answer is "15"

 

First A.P., a = 1, d = 4
S_2n = 2n22n2 [2 * 1+(2n −1)4]
S_2n = n(2 + 8n – 4) = n(8n – 2) = 8n² – 2n
For the second AP, a = 56, d = 2
S_n = n2n2 [2 * 56+(n −1)2]
S_n = n2n2 [112+2n −2] = 2n22n2 (110+2n)
Sum of 2n terms of AP {1, 5, 9, 13, …..} is greater than sum of n terms of A.P. = {56, 58, 60, …}
8n² – 2n > n2n2 (110+2n)
16n² – 4n > 110n + 2n2
14n² > 114n
7n > 57
n > 577577
The smallest value n can take = 9.

The question is "What is the smallest value n can take?"

Hence the answer is "9"

a, b, c and d are in A.P.
Dividing through out this A.P. by abcd, we get
aabcdaabcd , babcdbabcd , cabcdcabcd , dabcddabcd are in A.P.
1bcd1bcd , 1acd1acd , 1abd1abd , 1abc1abc are in A.P.
=> bcd, acd, abd and abc are in H.P.

The question is "a, b, c and d are in A.P., What can we say about terms bcd, acd, abd and abc?"

Hence the answer is "They are in H.P."

S_n - S_n = t_n
Substitute m instead of n
S_m - S_m-1 = t_m
We know that S_n = n³ + n² + n + 1

Hence m³ + m² + m + 1
m³ + m² + m + 1 - [(m-1)³ + (m-1)² + (m-1) + 1 ] = 291
m³ + m² + m + 1 - [m³ - (3m)² + 3m - 1 + m² - 2m + 1 + m - 1 + 1] = 291
1 + (3m)² + 3m + 1 -2m - 1 = 291
(-3m)² + m - 290 = 0
(3m)² - m + 290 = 0

Solving above equation we get m = -29, 30
M cannot be negative
Hence m = 30

The question is \"If S_n = n³ + n² + n + 1 , where S_n denotes the sum of the first n terms of a series and t_m = 291, then m is equal to?\"

Hence the answer is \"30\"

S_∞ = 12 and a = 8
S_∞ = a1−ra1−r 
12 = 81−r81−r 
1 - r = 2323 
r = 1313 
Now t₄ = 8 * 133133
t₄ = 827827

The question is "Sum of infinite terms of a GP is 12. If the first term is 8, what is the 4th term of this GP?"

Hence the answer is "8/27"

The series is in the form n(n + 1)² = n³ + 2n² + n

Then Σ n³ + 2n² + n = [n(n+1)2n(n+1)2 ]² + 2 [n (n + 1)2n+1)62n+1)6] + [n(n+1)2n(n+1)2]
Substituting n = 10 will give the sum of the series. Thus Σ [10∗11)210∗11)2]² + [10∗11∗21)610∗11∗21)6 ] + [10∗11)210∗11)2]

The question is "Find sum of 2 ² + 2 * 3² + 3 * 4² + 4 * 5².....10 * 11²?"

Hence the answer is "3850"

Let a₁, a₂ be the salaries of their first year and d₁ and d₂ be the difference between salaries of two consecutive years.
Now, the sum of p terms of their salaries are
S_p = p2p2 * [2a₁ + (p-1)d₁]
S'_p = p2p2 * [2a₂ + (p-1)d₂]
p2∗[2a1+(p−1)d1]p2∗[2a2+(p−1)d2]p2∗[2a1+(p−1)d1]p2∗[2a2+(p−1)d2] = 4p+12p+174p+12p+17
[2a1+(p−1)d1][2a2+(p−1)d2][2a1+(p−1)d1][2a2+(p−1)d2] = 4p+12p+174p+12p+17 -------------------- (1)
Now,
Ratio of Salaries earned by them in the 7^th year
= a1+6d1a2+6d2a1+6d1a2+6d2
= 2a1+12d12a2+12d22a1+12d12a2+12d2 (Multiplying and Dividing by 2)
= 2a1+(13−1)d1[2a2+(13−1)d22a1+(13−1)d1[2a2+(13−1)d2
= 4∗13+12∗13+174∗13+12∗13+17
= 53 : 43

The question is "find the ratio of amount earned by them in the 7^th year."

Hence the answer is "53 : 43"

For solving these type of questions we must always take the numbers as
(a-3d),(a-d),(a+d),(a+3d)
So, Let the investment made by Ram in these years be (a-3d),(a-d),(a+d),(a+3d)
Then,
(a-3d) + (a-d) + (a+d) +(a+3d) = 2000
=> 4a = 2000
=> a = 500

Now, again AQAQ
(a-3d)² +(a-d)² +(a+d)² +(a+3d)² = 1200000
=> 4(a² + 5 d²) = 1200000
=> (a² + 5 d²) = 300000
=> 250000 + 5d² = 300000
=> 5d² = 50000
=> d² = 10000
=> d = ±100

∴ d = 100 ( As the investment increases every year so d = -100 not possible)
Hence the investments made are 200,400,600,800

Note - If there are 3 number in series then take the numbers as (a-d),a,(a+d)
If there are 5 number in series then take the numbers as (a-2d),(a-d), a,(a+d),(a+2d)
…….. so on

While If there are 4 number in the series then take the numbers as (a-3d),(a-d), ,(a+d),(a+3d)
If there are 6 number in the series then take the numbers as (a-5d),(a-3d),(a-d), ,(a+d),(a+3d), (a+5d)
……………. so on

The question is \"Find the investment made by ram in each year respectively. It is also known that he always invest more than the previous year.\"

Hence the answer is \"200,400,600,800\"

.4 + .44 + .444+……. to n terms
= 4 * [0.1 +0.11 +0.111+……..to n terms]
= 4949 * [0.9 +0.99 + 0.999+…….to n terms] (Multiplying and dividing by 9)
= 4949 * [(1-0.1)+(1-0.01)+(1-0.001)+…….to n terms]
= 4949 * [(1+1+1…..to n terms)-(0.1+0.01+0.001…to n terms)]
= 4949 * [ n - 110∗(1−110n)1−110110∗(1−110n)1−110 ]
= 4949 * [n - 10n−19∗10n10n−19∗10n ]
= 4949 * [n - 1919 * (1 - 110n110n)]
= 481481 * [9n - (1 - 110n110n]
= 481481 * [9n - 1 + 110n110n]

The question is "Find the sum of the series .4 + .44 + .444……. to n terms"

Hence the answer is "4/ 81 [9n-1+1/10n]"

Since p,q,r are in G.P. we have q² = pr. On solving px² +2qx +r = 0 we get
x = −2q±√(4q2−4pr)2p−2q±√(4q2−4pr)2p
=> x = −2q2p−2q2p (q² = pr)
=> x = −qp−qp
Thus x = −qp−qp is the repeated root of px²+2qx+r = 0
∴ x = −qp−qp is also a root of dx²+2ex+f = 0
=> d(−qp−qp)² + 2e(−qp−qp) + f = 0
=> dq2−2eqp+fp2p2dq2−2eqp+fp2p2 = 0
=> dq² - 2eqp + fp² = 0
=> dpdp - 2 * eqeq + fpq2fpq2 = 0 [On dividing by pq²]
=> dpdp - 2 * eqeq + frfr = 0
=> dpdp + frfr = 2 * eqeq
Hence it can be clearly seen that dpdp , eqeq , frfr are in A.P

The question is "If the equation px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then in which type of progression is 

Hence the answer is are in A.P."

Let the numbers be x-a, x, x+a
∴ For last 3 terms to be in G.P:-
(x+a)² = x(x-a+40)
x²+a²+2ax = x²-ax+40x
a² = -3ax + 40x
=> x = a2(40−3a)a2(40−3a)
Now, since x is a positive integer, putting different integral value of a .
Only a = 12 gives a integer value of x = 36

(Note, a=b also give integral value x=b but then the first term becomes 0)
∴ x = 36, a = 12
∴ x – a = 24
x = 36
x+a = 48
x-a+40 = 64
∴ Sum of all the terms = 64+48+36+24 = 172
The terms are - 24,36,48,64

The question is "Find the sum of all the terms, If the first 3 terms among 4 positive integers are in A.P and the last 3 terms are in G.P."

Hence the answer is "172"