Quiz 5

Given, Second term of an AP is 8 => a+d =8, 8th term is 2 more than thrice the second term => a+7d = 2 +3(a+d) = 2+3*8 =26 .
a+d =8 -------------- 1
a+7d = 26 -------------- 2
Solving for d in the two equations, we get d = 3 and a =5.
Sum of n terms in an AP = n2n2 * [2a +(n-1)d].
=> Sum upto 8 terms in this AP = 8282 * [2*5 + (8-1)3] => 4*[10+21] = 4*31 = 124.

The question is "Second term in an AP is 8 and the 8th term is 2 more than thrice the second term. Find the sum up to 8 terms of this AP."

Hence the answer is "124"

Sum of first 12 terms is equal to sum of first 14 terms.
Sum of first 14 terms = Sum of first 12 terms + 13^th term + 14^th term
=> 13^th term + 14^th term = 0

Let us assume 13^th term = k, common ratio = r. 14^th term will be kr.
k + kr = 0
k (1 + r) = 0
=> r = -1 as k cannot be zero

Common ratio = -1.
Now, if the first term of this GP is a, second term would be -a, third would be a and so on
The GP would be a, -a, a, -a, a, -a,...

Sum to even number of terms = 0
Sum to odd number of terms = a

Sum to 17 terms is 92 => a = 92
Third term = a = 92

The question is "what is the third term in the GP?"

Hence the answer is "92"

Choice A is the correct answer.

|t₈|=|t₁₆|. This can happen under two scenarios t₈ = t₁₆ or t₈ = – t₁₆.
If t₈ = t₁₆, the common difference would be 0 suggesting that t₃ would be equal to t₇.
However, we know t₃ is not equal to t₇, so the common difference cannot be zero.
This tells us that t₈ = – t₁₆ Or, t₈ + t₁₆ = 0.
If t₈ + t₁₆ = 0, then t₁₂ = 0. t₁₂ = t₈ + 4d, and t₁₆ – 4d So, t₁₂ = t8+t162t8+t162
For any two terms in an AP, the mean is the term right in between them.
So, t₁₆ is the arithmetic mean of t₈ and t₁₆.
So, t₁₂ = 0.
Now, S₂₃ = 23 * t₁₂. We know that average of n terms in an A.P. is the middle term.
This implies that sum of n terms in an A.P., is n times the middle term. So, S₂₃ = 0.

The question is "If |t₈| = |t₁₆| and t₃ is not equal to t₇, what is S₂₃?"

Hence the answer is "zero."

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"

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"

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"